Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations

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Big Ideas Math Book Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations

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Solving Quadratic Equations Maintaining Mathematical Proficiency

Factor the trinomial.
Question 1.
x2 + 10x + 25
Answer:
The given equation is:
x² + 10x + 25
Write the given equation in the form of
a² + 2ab + b²
We know that,
a² + 2ab + b² = (a + b )²
So,
x² + 10x + 25
= x² + 2 (5) (x) + 5²
Compare the above equation with
a² + 2ab + b²
So,
a = x and b = 5
So,
x² + 2 (5) (x) + 5² = (x + 5)²
Hence,
The factor form of the given trinomial is: (x + 5)²

Question 2.
x2 – 20x + 100
Answer:
The given equation is:
x² – 20x + 100
Write the given equation in the form of
a² – 2ab + b²
We know that,
a² – 2ab + b² = (a – b )²
So,
x² – 20x + 100
= x² – 2 (10) (x) + 10²
Compare the above equation with
a² – 2ab + b²
So,
a = x and b = 10
So,
x² – 2 (10) (x) + 10² = (x – 10)²
Hence,
The factor form of the given trinomial is: (x – 10)²

Question 3.
x2 + 12x + 36
Answer:
The given equation is:
x² + 12x + 36
Write the given equation in the form of
a² + 2ab + b²
We know that,
a² + 2ab + b² = (a + b )²
So,
x² + 12x + 36
= x² + 2 (6) (x) + 6²
Compare the above equation with
a² + 2ab + b²
So,
a = x and b = 6
So,
x² + 2 (6) (x) + 6² = (x + 6)²
Hence,
The factor form of the given trinomial is: (x + 6)²

Question 4.
x2 – 18x + 81
Answer:
The given equation is:
x² – 18x + 81
Write the given equation in the form of
a² – 2ab + b²
We know that,
a² – 2ab + b² = (a – b )²
So,
x² – 18x + 81
= x² – 2 (9) (x) + 9²
Compare the above equation with
a² – 2ab + b²
So,
a = x and b = 9
So,
x² – 2 (9) (x) + 9² = (x – 9)²
Hence,
The factor form of the given trinomial is: (x – 9)²

Question 5.
x2 + 16x + 64
Answer:
The given equation is:
x² + 16x + 64
Write the given equation in the form of
a² + 2ab + b²
We know that,
a² + 2ab + b² = (a + b )²
So,
x² + 16x + 64
= x² + 2 (8) (x) + 8²
Compare the above equation with
a² + 2ab + b²
So,
a = x and b = 8
So,
x² + 2 (8) (x) + 8² = (x + 8)²
Hence,
The factor form of the given trinomial is: (x + 8)²

Question 6.
x2 – 30x + 225
Answer:
The given equation is:
x² – 30x + 225
Write the given equation in the form of
a² – 2ab + b²
We know that,
a² – 2ab + b² = (a – b )²
So,
x² – 30x + 225
= x² – 2 (15) (x) + 15²
Compare the above equation with
a² – 2ab + b²
So,
a = x and b = 15
So,
x² – 2 (15) (x) + 15² = (x – 15)²
Hence,
The factored form of the given trinomial is: (x – 15)²

Solve the system of linear equations by graphing.
Question 7.
y = -5x + 3
y = 2x – 4
Answer:
The given system of linear equations are:
y = -5x + 3
y = 2x – 4
So,
The representation of the system of linear equations in the coordinate plane is:

From the graph,
We can observe that the intersection point of the given system of linear equations is: (1, -2)
Hence, from the above,
We can conclude that the solution of the given system of linear equations is: (1, -2)

Question 8.
y = \(\frac{3}{2}\)x – 2
y = –\(\frac{1}{4}\)x + 5
Answer:
The given system of linear equations are:
y = \(\frac{3}{2}\)x – 2
y = –\(\frac{1}{4}\)x + 5
So,
The representation of the system of linear equations in the coordinate plane is:

From the graph,
We can observe that the intersection point of the given system of linear equations is: (4, 4)
Hence, from the above,
We can conclude that the solution of the given system of linear equations is: (4, 4)

Question 9.
y = \(\frac{1}{2}\)x + 4
y = -3x – 3
Answer:
The given system of linear equations are:
y = \(\frac{1}{2}\)x + 4
y = -3x – 3
So,
The representation of the system of linear equations in the coordinate plane is:

From the graph,
We can observe that the intersection point of the given system of linear equations is: (-2, 3)
Hence, from the above,
We can conclude that the solution of the given system of linear equations is: (-2, 3)

Question 10.
ABSTRACT REASONING
What value of c makes x2 + bx + c a perfect square trinomial?
Answer:
The given linear equation is:
x² + bx + c
We know that,
To make the given linear equation a perfect square trinomial,
The value of c must be a square
Fora perfect square trinomial,
We compare the equation with
a² + 2ab + b²
So,
x² + 2 (\(\frac{b}{2}\)) (x) + c
By comparing the equation with the given equation,
We get,
c = \(\frac{b}{2}\)
Hence, from the above,
We can conclude that the value of c to make the given linear equation a perfect square trinomial is:
\(\frac{b}{2}\)

Solving Quadratic Equations Mathematical Practices

Mathematically proficient students monitor their work and change course as needed.

Monitoring Progress

Question 1.
Use the graph in Example 1 to approximate the negative solution of the equation x2 + x – 1 = 0 to the nearest thousandth.

Answer:
The given equation is:
x² + x – 1 = 0
From the graph,
We can estimate the approximate solution of the given equation is: 0.65
So,
To find the negative solution of the given equation,
The solution of the given equation will change from positive to negative
Now,
For x = 0.65,
The approximate value of the given equation is:
(0.65)² + 0.65 – 1 = 0.0725
For x = 0.64,
The approximate value of the given equation is:
(0.64)² + 0.64 – 1 = 0.0496
For x = 0.63,
The approximate value of the given equation is:
(0.63)² + 0.63 – 1 = 0.0269
For x = 0.62,
The approximate value of the given equation is:
(0.62)² + 0.62 – 1 = 0.0044
For x = 0.61,
The approximate value of the given equation is:
(0.61)² + 0.61 – 1 = -0.0179
For x = 0.60,
The approximate value of the given equation is:
(0.60)² + 0.60 – 1 = -0.04
For x = 0.59,
The approximate value of the given equation is:
(0.59)² + 0.59 – 1 = -0.0619
Hence, from the above,
We can conclude that the negative solution of the given equation is: 0.61

Question 2.
The graph of y = x2 + x – 3 is shown. Approximate both solutions of the equation x2 + x – 3 = 0 to the nearest thousandth.
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 1
Answer:
The given equation is:
y = x² + x – 3
From the graph,
We can estimate
The approximate positive solution of the given equation is: 1.60
The approximate negative solution of the given equation is: 2.2
Now,
For the positive solution of the given equation:
For x = 1.60,
The approximate value of the given equation is:
(1.60)² + 1.60 – 3 = 1.162
For x = 1.59,
The approximate value of the given equation is:
(1.59)² + 1.59 – 3 = 1.118
For x = 1.30,
The approximate value of the given equation is:
(1.30)² + 1.30 – 3 = -0.01
Hence, from the above,
We can conclude that the positive solution of the given equation is: 1.30
Now,
For the negative solution of the given equation:
For x = 2.20,
The approximate value of the given equation is:
(2.20)² + 2.20 – 3 = 4.040
For x = 2.10,
The approximate value of the given equation is:
(2.10)² + 2.10 – 3 = 3.510
For x = 1.90,
The approximate value of the given equation is:
(1.90)² + 1.90 – 3 = 2.510
For x = 1.20,
The approximate value of the given equation is:
(1.20)² + 1.20 – 3 = -0.360
Hence, from the above,
We can conclude that the negative solution of the given equation is: 1.90

Lesson 9.1 Properties of Radicals

Essential Question
How can you multiply and divide square roots?
Answer:
We can multiply and divide square roots by using the “Product Rule” and “Quotient Rule”
The product rule is:
\(\sqrt{a}\) ⋅ \(\sqrt{b}\) = \(\sqrt{{a}\cdot b}\)
The quotient rule is:
\(\frac{\sqrt{a}}{\sqrt{b}}\) = \(\sqrt{\frac{a}{b}}\)

EXPLORATION 1

Operations with Square Roots
Work with a partner. For each operation with square roots, compare the results obtained using the two indicated orders of operations. What can you conclude?
a. Square Roots and Addition
Is \(\sqrt{36}\) + \(\sqrt{64}\) equal to \(\sqrt{36+64}\)?
In general, is \(\sqrt{a}\) + \(\sqrt{b}\) equal to \(\sqrt{a+b}\)? Explain your reasoning.
Answer:
The given expressions are:
\(\sqrt{36}\) + \(\sqrt{64}\)
\(\sqrt{36 + 64}\)
Now,
Compare the given expressions with
\(\sqrt{a}\) + \(\sqrt{b}\) and \(\sqrt{a + b}\)
We can get,
a = 36 and b = 64
Now,
\(\sqrt{36}\) + \(\sqrt{64}\)
= 6 + 8
= 14
\(\sqrt{36 + 64}\)
= \(\sqrt{100}\)
= 10
Hence, from the above,
We can conclude that
\(\sqrt{a}\) + \(\sqrt{b}\) is not equal to \(\sqrt{a + b}\)

b. Square Roots and Multiplications
\(\sqrt{4}\) • \(\sqrt{9}\) equal to \(\sqrt{{4} \cdot 9}\)?
In general, is \(\sqrt{a}\) • \(\sqrt{b}\) equal to \(\sqrt{{a} \cdot b}\)? Explain your reasoning.
Answer:
The given expressions are:
\(\sqrt{4}\) ⋅ \(\sqrt{9}\)
\(\sqrt{{4}\cdot9}\)
Now,
Compare the given expressions with
\(\sqrt{a}\) ⋅ \(\sqrt{b}\) and \(\sqrt{{a}\cdot b}\)
We can get,
a = 4 and b = 9
Now,
\(\sqrt{4}\) ⋅ \(\sqrt{9}\)
= 2 ⋅ 3
= 6
\(\sqrt{{4}\cdot 9}\)
= \(\sqrt{36}\)
= 6
Hence, from the above,
We can conclude that
\(\sqrt{a}\) ⋅ \(\sqrt{b}\) is  equal to \(\sqrt{{a}\cdot b}\)

c. Is \(\sqrt{64}\) – \(\sqrt{36}\) equal to \(\sqrt{64 – 36}\)?
In general, is \(\sqrt{a}\) – \(\sqrt{b}\) equal to \(\sqrt{a-b}\)? Explain your reasoning.
Answer:
The given expressions are:
\(\sqrt{64}\) – \(\sqrt{36}\)
\(\sqrt{64 – 36}\)
Now,
Compare the given expressions with
\(\sqrt{a}\) – \(\sqrt{b}\) and \(\sqrt{a – b}\)
We can get,
a = 64 and b = 36
Now,
\(\sqrt{64}\) – \(\sqrt{36}\)
= 8 – 6
= 2
\(\sqrt{64 – 36}\)
= \(\sqrt{28}\)
= 5.291
Hence, from the above,
We can conclude that
\(\sqrt{a}\) – \(\sqrt{b}\) is not equal to \(\sqrt{a – b}\)

d. Square Roots and Division
Is \(\frac{\sqrt{100}}{\sqrt{4}}\) equal to \(\sqrt{\frac{100}{4}}\)?
In general, is \(\frac{\sqrt{a}}{\sqrt{b}}\) equal to \(\sqrt{\frac{a}{b}}\)? Explain your reasoning.
Answer:
The given expressions are:
\(\frac{\sqrt{100}}{\sqrt{4}}\)
\(\sqrt{\frac{100}{4}}\)
Now,
Compare the given expressions with
\(\frac{\sqrt{a}}{\sqrt{b}}\)  and \(\sqrt{\frac{a}{b}}\)
We can get,
a = 100 and b = 4
Now,
\(\frac{\sqrt{100}}{\sqrt{4}}\)
= \(\frac{10}{2}\)
= 5
\(\sqrt{\frac{100}{4}}\)
= \(\sqrt{25}\)
= 5
Hence, from the above,
We can conclude that
\(\frac{\sqrt{a}}{\sqrt{b}}\)  is equal to \(\sqrt{\frac{a}{b}}\)

EXPLORATION 2

Writing Counterexamples
Work with a partner. A counterexample is an example that proves that a general statement is not true. For each general statement in Exploration 1 that is not true, write a counterexample different from the example given.
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 1

Communicate Your Answer

Question 3.
How can you multiply and divide square roots?
Answer:
We can multiply and divide square roots by using the “Product Rule” and “Quotient Rule”
The product rule is:
\(\sqrt{a}\) ⋅ \(\sqrt{b}\) = \(\sqrt{{a}\cdot b}\)
The quotient rule is:
\(\frac{\sqrt{a}}{\sqrt{b}}\) = \(\sqrt{\frac{a}{b}}\)

Question 4.
Give an example of multiplying square roots and an example of dividing square roots that are different from the examples in Exploration 1.
Answer:
Example of multiplying square roots:
We know that,
We use the “Product Rule” for the multiplication of square roots
The product rule is:
\(\sqrt{a}\) ⋅ \(\sqrt{b}\) = \(\sqrt{{a}\cdot b}\)
Now,
Let,
The values of a nad b are:
a = 144 and b = 36
So,
\(\sqrt{144}\) ⋅ \(\sqrt{36}\)
= 12 ⋅ 6
= 72
\(\sqrt{{144}\cdot 36}\)
= \(\sqrt{5,184}\)
= 72
Example of dividing square roots:
We know that,
We use the “Quotient Rule” for the division of square roots
The quotient rule is:
\(\frac{\sqrt{a}}{\sqrt{b}}\) = \(\sqrt{\frac{a}{b}}\)
Now,
Let,
The values of a and b are:
a = 144 and b = 36
So,
\(\frac{\sqrt{144}}{\sqrt{36}}\)
= \(\frac{12}{6}\)
= 2
\(\sqrt{\frac{144}{36}}\)
= \(\sqrt{\frac{12}{3}}\)
= \(\sqrt{4}\)
= 2

Question 5.
Write an algebraic rule for each operation.
a. the product of square roots
b. the quotient of square roots
Answer:
a.
The algebraic rule for the product of square roots:
\(\sqrt{a}\) ⋅ \(\sqrt{b}\) = \(\sqrt{{a}\cdot b}\)
The algebraic rule for the division of square roots is:
\(\frac{\sqrt{a}}{\sqrt{b}}\) = \(\sqrt{\frac{a}{b}}\)

9.1 Exercise

Monitoring Progress

Simplify the expression.
Question 1.
\(\sqrt{24}\)
Answer:
The given expression is:
\(\sqrt{24}\)
We know that,
\(\sqrt{{a}\cdot b}\) = \(\sqrt{a}\) ⋅ \(\sqrt{b}\)
So,
\(\sqrt{24}\)
= \(\sqrt{{4}\cdot 6}\)
= \(\sqrt{4}\) ⋅ \(\sqrt{6}\)
= 2 \(\sqrt{6}\)
Hence, from the above,
We can conclude that
\(\sqrt{24}\) = 2 \(\sqrt{6}\)

Question 2.
–\(\sqrt{80}\)
Answer:
The given expression is:
–\(\sqrt{80}\)
We know that,
\(\sqrt{{a}\cdot b}\) = \(\sqrt{a}\) ⋅ \(\sqrt{b}\)
So,
–\(\sqrt{80}\)
= –\(\sqrt{{16}\cdot 5}\)
= –\(\sqrt{16}\) ⋅ \(\sqrt{5}\)
= -4 \(\sqrt{5}\)
Hence, from the above,
We can conclude that
–\(\sqrt{80}\) = -4 \(\sqrt{5}\)

Question 3.
\(\sqrt{49 x^{3}}\)
Answer:
The given expression is:
\(\sqrt{49 x^{3}}\)
We know that,
\(\sqrt{{a}\cdot b}\) = \(\sqrt{a}\) ⋅ \(\sqrt{b}\)
So,
\(\sqrt{49 x^{3}}\)
= \(\sqrt{{49 x^{2}}\cdot x}\)
= \(\sqrt{49 x^{2}}\) ⋅ \(\sqrt{x}\)
= 7x \(\sqrt{x}\)
Hence, from the above,
We can conclude that
\(\sqrt{49 x^{3}}\) = 7x \(\sqrt{x}\)

Question 4.
\(\sqrt{49 n^{5}}\)
Answer:
The given expression is:
\(\sqrt{49 n^{5}}\)
We know that,
\(\sqrt{{a}\cdot b}\) = \(\sqrt{a}\) ⋅ \(\sqrt{b}\)
So,
\(\sqrt{49 n^{5}}\)
= \(\sqrt{{49 n^{4}}\cdot n}\)
= \(\sqrt{49 n^{4}}\) ⋅ \(\sqrt{n}\)
= 7n² \(\sqrt{n}\)
Hence, from the above,
We can conclude that
\(\sqrt{49 n^{5}}\) = 7n²  \(\sqrt{n}\)

Simplify the expression.
Question 5.
\(\sqrt{\frac{23}{9}}\)
Answer:
The given expression is:
\(\sqrt{\frac{23}{9}}\)
We know that,
\(\frac{\sqrt{a}}{\sqrt{b}}\) = \(\sqrt{\frac{a}{b}}\)
So,
\(\sqrt{\frac{23}{9}}\)
= \(\sqrt{\frac{23}{9}}\)
= \(\frac{\sqrt{23}}{\sqrt{9}}\)
= \(\frac{\sqrt{23}}{3}\)
Hence, from the above,
We can conclude that
\(\sqrt{\frac{23}{9}}\) = \(\frac{\sqrt{23}}{3}\)

Question 6.
–\(\sqrt{\frac{17}{100}}\)
Answer:
The given expression is:
–\(\sqrt{\frac{17}{100}}\)
We know that,
\(\frac{\sqrt{a}}{\sqrt{b}}\) = \(\sqrt{\frac{a}{b}}\)
So,
–\(\sqrt{\frac{17}{100}}\)
= –\(\sqrt{\frac{17}{100}}\)
= –\(\frac{\sqrt{17}}{\sqrt{100}}\)
= –\(\frac{\sqrt{17}}{10}\)
Hence, from the above,
We can conclude that
–\(\sqrt{\frac{17}{100}}\) = –\(\frac{\sqrt{17}}{10}\)

Question 7.
\(\sqrt{\frac{36}{z^{2}}}\)
Answer:
The given expression is:
\(\sqrt{\frac{36}{z^{2}}}\)
We know that,
\(\frac{\sqrt{a}}{\sqrt{b}}\) = \(\sqrt{\frac{a}{b}}\)
So,
\(\sqrt{\frac{36}{z^{2}}}\)
= \(\sqrt{\frac{36}{z^{2}}}\)
= \(\frac{\sqrt{36}}{\sqrt{z^{2}}}\)
= \(\frac{6}{z}\)
Hence, from the above,
We can conclude that
\(\sqrt{\frac{36}{z^{2}}}\) = \(\frac{6}{z}\)

Question 8.
\(\sqrt{\frac{4 x^{2}}{64}}\)
Answer:
The given expression is:
\(\sqrt{\frac{4 x^{2}}{64}}\)
We know that,
\(\frac{\sqrt{a}}{\sqrt{b}}\) = \(\sqrt{\frac{a}{b}}\)
So,
\(\sqrt{\frac{4 x^{2}}{64}}\)
= \(\sqrt{\frac{4 x^{2}}{64}}\)
= \(\frac{\sqrt{4 x^{2}}}{\sqrt{64}}\)
= \(\frac{2x}{8}\)
= \(\frac{x}{4}\)
Hence, from the above,
We can conclude that
\(\sqrt{\frac{4 x^{2}}{64}}\) = \(\frac{x}{4}\)

Question 9.
\(\sqrt[3]{54}\)
Answer:
The given expression is:
\(\sqrt[3]{54}\)
We know that,
\(\sqrt[3]{a}{b}\) = \(\sqrt[3]{a}\) ⋅ \(\sqrt[3]{b}\)
So,
\(\sqrt[3]{54}\)
= \(\sqrt[3]{54}{1}\)
= \(\sqrt[3]{54}\) ⋅ \(\sqrt[3]{1}\)
= \(\sqrt[3]{27 × 2}\)
= \(\sqrt[3]{27}\) ⋅ \(\sqrt[3]{2}\)
= 3\(\sqrt[3]{2}\)
Hence, from the above,
We can conclude that
\(\sqrt[3]{54}\) = 3\(\sqrt[3]{2}\)

Question 10.
\(\sqrt[3]{16 x^{4}}\)
Answer:
The given expression is:
\(\sqrt[3]{16 x^{4}}\)
We know that,
\(\sqrt[3]{a}{b}\) = \(\sqrt[3]{a}\) ⋅ \(\sqrt[3]{b}\)
So,
\(\sqrt[3]{16 x^{4}}\)
= \(\sqrt[3]{x^{3}}{16x}\)
= \(\sqrt[3]{x^{3}}\) ⋅ \(\sqrt[3]{16x}\)
= x\(\sqrt[3]{16x}\)
Hence, from the above,
We can conclude that
\(\sqrt[3]{16 x^{4}}\) = x\(\sqrt[3]{16x}\)

Question 11.
\(\sqrt[3]{\frac{a}{-27}}\)
Answer:
The given expression is:
\(\sqrt[3]{\frac{a}{-27}}\)
We know that,
\(\sqrt[3]{\frac{a}{b}}\) = \(\sqrt[3]{a}\) / \(\sqrt[3]{b}\)
So,
\(\sqrt[3]{\frac{a}{-27}}\)
= \(\sqrt[3]{a}\) / \(\sqrt[3]{-27}\)
=\(\sqrt[3]{a}\) / \(\sqrt[3]{-3^{3}}\)
= –\(\sqrt[3]{a}\) / 3
Hence, from the above,
We can conclude that
\(\sqrt[3]{\frac{a}{-27}}\) = –\(\sqrt[3]{a}\) / 3

Question 12.
\(\sqrt[3]{\frac{25 c^{7} d^{3}}{64}}\)
Answer:
The given expression is:
\(\sqrt[3]{\frac{25 c^{7} d^{3}}{64}}\)
We know that,
\(\sqrt[3]{\frac{a}{b}}\) = \(\sqrt[3]{a}\) / \(\sqrt[3]{b}\)
So,
\(\sqrt[3]{\frac{25 c^{7} d^{3}}{64}}\)
= \(\sqrt[3]{25 c^{7} d^{3}}\) / \(\sqrt[3]{64}\)
=\(\sqrt[3]{25 c^{3} c^{4} d^{3}}\) / \(\sqrt[3]{4^{3}}\)
= c²d\(\sqrt[3]{25c}\) / 4
Hence, from the above,
We can conclude that
\(\sqrt[3]{\frac{25 c^{7} d^{3}}{64}}\) = c²d\(\sqrt[3]{25c}\) / 4

Simplify the expression.
Question 13.
\(\frac{1}{\sqrt{5}}\)
Answer:
The given expression is:
\(\frac{1}{\sqrt{5}}\)
We have to rationalize the denominator to simplify the given expression
So,
To rationalize the denominator, multiply and divide the given fraction with \(\sqrt{5}\)
So,
\(\frac{1}{\sqrt{5}}\)
= \(\frac{1}{\sqrt{5}}\) × (\(\sqrt{5}\) / \(\sqrt{5}\))
= \(\sqrt{5}\) / (\(\sqrt{5}\) × \(\sqrt{5}\))
= \(\sqrt{5}\) / 5
Hence, from the above,
We can conclude that
\(\frac{1}{\sqrt{5}}\) = \(\sqrt{5}\) / 5

Question 14.
\(\frac{\sqrt{10}}{\sqrt{3}}\)
Answer:
The given expression is:
\(\frac{\sqrt{10}}{\sqrt{3}}\)
We have to rationalize the denominator to simplify the given expression
So,
To rationalize the denominator, multiply and divide the given fraction with \(\sqrt{3}\)
So,
\(\frac{\sqrt{10}}{\sqrt{3}}\)
= \(\frac{\sqrt{10}}{\sqrt{3}}\) × (\(\sqrt{3}\) / \(\sqrt{3}\))
= \(\sqrt{10 × 3}\) / (\(\sqrt{3}\) × \(\sqrt{3}\))
= \(\sqrt{30}\) / 3
Hence, from the above,
We can conclude that
\(\frac{\sqrt{10}}{\sqrt{3}}\) = \(\sqrt{30}\) / 3

Question 15.
\(\frac{7}{\sqrt{2 x}}\)
Answer:
The given expression is:
\(\frac{7}{\sqrt{2 x}}\)
We have to rationalize the denominator to simplify the given expression
So,
To rationalize the denominator, multiply and divide the given fraction with \(\sqrt{2 x}\)
So,
\(\frac{7}{\sqrt{2 x}}\)
= \(\frac{7}{\sqrt{2 x}}\) × (\(\sqrt{2 x}\) / \(\sqrt{2 x}\))
= 7\(\sqrt{2 x}\) / (\(\sqrt{2 x}\) × \(\sqrt{2 x}\))
= 7\(\sqrt{2 x}\) / 2x
Hence, from the above,
We can conclude that
\(\frac{7}{\sqrt{2 x}}\) = 7\(\sqrt{2 x}\) / 2x

Question 16.
\(\sqrt{\frac{2 y^{2}}{3}}\)
Answer:
The given expression is:
\(\sqrt{\frac{2 y^{2}}{3}}\)
We know that,
\(\sqrt{\frac{a}{b}}\) = \(\frac{\sqrt{a}}{\sqrt{b}}\)
So,
\(\sqrt{\frac{2 y^{2}}{3}}\)
= \(\frac{\sqrt{2 y^{2}}}{\sqrt{3}}\)
We have to rationalize the denominator to simplify the given expression
So,
To rationalize the denominator, multiply and divide the given fraction with \(\sqrt{3}\)
So,
\(\frac{\sqrt{2 y^{2}}}{\sqrt{3}}\)
= \(\frac{\sqrt{2 y^{2}}}{\sqrt{3}}\) × (\(\sqrt{3}\) / \(\sqrt{3}\))
= \(\sqrt{2 y^{2} × 3}\) / (\(\sqrt{3}\) × \(\sqrt{3}\))
= y\(\sqrt{6}\) / 3
Hence, from the above,
We can conclude that
\(\sqrt{\frac{2 y^{2}}{3}}\) = y\(\sqrt{6}\) / 3

Question 17.
\(\frac{5}{\sqrt[3]{32}}\)
Answer:
The given expression is:
\(\frac{5}{\sqrt[3]{32}}\)
We have to rationalize the denominator to simplify the given expression
So,
To rationalize the denominator, multiply and divide the given fraction with \(\sqrt[3]{2}\)
So,
\(\frac{5}{\sqrt[3]{32}}\)
= \(\frac{5}{\sqrt[3]{32}}\) × (\(\sqrt[3]{2}\) / \(\sqrt[3]{2}\))
= 5\(\sqrt[3]{2}\) / (\(\sqrt[3]{32}\) × \(\sqrt[3]{2}\))
= 5\(\sqrt[3]{2}\) / 4
Hence, from the above,
We can conclude that
\(\frac{5}{\sqrt[3]{32}}\) = 5\(\sqrt[3]{2}\) / 4

Question 18.
\(\frac{8}{1+\sqrt{3}}\)
Answer:
The given expression is:
\(\frac{8}{1 + \sqrt{3}}\)
We have to rationalize the denominator to simplify the given expression
So,
To rationalize the denominator, multiply and divide the given fraction with \(1 – \sqrt{3}\)
So,
\(\frac{8}{1 + \sqrt{3}}\)
= \(\frac{8}{1 + \sqrt{3}}\) × (\(1 – \sqrt{3}\) / \(1 – \sqrt{3}\))
= 8\((1 – \sqrt{3})\) / (\(1 + \sqrt{3}\) × \(1 – \sqrt{3}\))
= 8\((1 – \sqrt{3})\) / -2
= – 8\((1 – \sqrt{3})\) / 2
Hence, from the above,
We can conclude that
\(\frac{8}{1 + \sqrt{3}}\) = -8 \((1 – \sqrt{3})\) / 2

Question 19.
\(\frac{\sqrt{13}}{\sqrt{5}-2}\)
Answer:
The given expression is:
\(\frac{\sqrt{13}}{\sqrt{5}-2}\)
We have to rationalize the denominator to simplify the given expression
So,
To rationalize the denominator, multiply and divide the given fraction with \(\sqrt{5} + 2\)
So,
\(\frac{\sqrt{13}}{\sqrt{5}-2}\)
= \(\frac{\sqrt{13}}{\sqrt{5}-2}\) × (\( \sqrt{5} – 2\) / \( \sqrt{5} – 2\))
= 13(\(\sqrt{{5}} – 2\)) / (\( \sqrt{5} – 2\) × \( \sqrt{5} + 2\))
= 13(\(\sqrt{{5}} – 2\)) / -1
= – 13(\(\sqrt{{5}} – 2\))
Hence, from the above,
We can conclude that
\(\frac{\sqrt{13}}{\sqrt{5}-2}\) = – 13(\(\sqrt{{5}} – 2\))

Question 20.
\(\frac{12}{\sqrt{2}+\sqrt{7}}\)
Answer:
The given expression is:
\(\frac{12}{\sqrt{2}+\sqrt{7}}\)
We have to rationalize the denominator to simplify the given expression
So,
To rationalize the denominator, multiply and divide the given fraction with \(\sqrt{2} -\sqrt{7} \)
So,
\(\frac{12}{\sqrt{2}+\sqrt{7}}\)
= \(\frac{12}{\sqrt{2}+\sqrt{7}}\) × (\(\sqrt{2} -\sqrt{7} \) / \(\sqrt{2} -\sqrt{7} \))
= 12(\(\sqrt{2} -\sqrt{7} \)) / (\(\sqrt{2} +\sqrt{7} \) × \(\sqrt{2} -\sqrt{7} \))
= 12(\(\sqrt{2} -\sqrt{7}\)) / -5
= – 12(\(\sqrt{2} -\sqrt{7} \))/ 5
Hence, from the above,
We can conclude that
\(\frac{12}{\sqrt{2}+\sqrt{7}}\) = – 12(\(\sqrt{2} -\sqrt{7} \))/ 5

Question 21.
WHAT IF?
In Example 6, how far can you see when your eye level is 35 feet above the water?
Answer:
In Example 6,
It is given that
d = \(\sqrt{\frac{3h}{2}}\)
Where,
‘h’ is the height that your eye can see above the water
‘d’ is the distance your eye can see
It is given that the value of ‘h’ is: 35 feet
So,
d = \(\sqrt{\frac{3 ×105}{2}}\)
d = \(\sqrt{\frac{315}{2}}\)
We know that,
\(\sqrt{\frac{a}{b}}\) = \(\frac{\sqrt{a}}{\sqrt{b}}\)
So,
\(\sqrt{\frac{315}{2}}\)
= \(\frac{\sqrt{315}}{\sqrt{2}}\)
We have to rationalize the denominator to simplify the given expression
So,
To rationalize the denominator, multiply and divide the given fraction with \(\sqrt{2}\)
So,
\(\sqrt{\frac{315}{2}}\)
= \(\frac{\sqrt{315}}{\sqrt{2}}\) × (\(\sqrt{2}\) / \(\sqrt{2}\))
= \(\sqrt{315 × 2}\) / (\(\sqrt{2}\) × \(\sqrt{2}\))
= \(\sqrt{630}\) / 2
= 12.54 feet
Hence, from the above,
We can conclude that the distance you can see when your eye level is 35 feet above the water is: 12.54 feet

Question 22.
The dimensions of a dance floor form a golden rectangle. The ratio of the length to the width of the golden rectangle is \(1 + \sqrt{5}\) : 2. The longer side of the dance floor is 50 feet. What is the length of the shorter side of the dance floor?
Answer:
It is given that the ratio of the length to the width of the golden rectangle is \(1 + \sqrt{5}\): 2 and the longer side of the dance floor is 50 feet.
Let the shorter side of the golden rectangle be ‘h’
So,
\(\frac{1+\sqrt{5}}{2}\) = \(\frac{50}{h}\)
h (\(1 + \sqrt{5}\)) = 50 × 2
h (\(1 + \sqrt{5}\)) = 100
h = \(\frac{100}{1 + \sqrt{5}}\)
We have to rationalize the denominator to simplify the given expression
So,
To rationalize the denominator, multiply and divide the given fraction with \(1 – \sqrt{5}\)
So,
\(\frac{100}{1 + \sqrt{5}}\)
= \(\frac{100}{1 + \sqrt{5}}\) × (\(1 – \sqrt{5}\) / \(1 – \sqrt{5}\))
= 100\((1 – \sqrt{5})\) / (\(1 + \sqrt{5}\) × \(1 – \sqrt{5}\))
= 100\((1 – \sqrt{5})\) / -4
= – 100\((1 – \sqrt{5})\) / 4
= 30.90 feet
Hence, from the above,
We can conclude that the length of the shorter side is 30.90 feet

Simplify the expression.
Question 23.
3\(\sqrt{2}\) – \(\sqrt{6}\) + 10\(\sqrt{2}\)
Answer:
The given expression is:
3\(\sqrt{2}\) – \(\sqrt{6}\) + 10\(\sqrt{2}\)
Now,
3\(\sqrt{2}\) – \(\sqrt{6}\) + 10\(\sqrt{2}\)
= \(\sqrt{2}\) (3 + 10) – \(\sqrt{6}\)
= \(\sqrt{2}\) (13) – \(\sqrt{6}\)
= 13\(\sqrt{2}\) – \(\sqrt{6}\)
Hence, from the above,
We can conclude that
3\(\sqrt{2}\) – \(\sqrt{6}\) + 10\(\sqrt{2}\) = 13\(\sqrt{2}\) – \(\sqrt{6}\)

Question 24.
4\(\sqrt{7}\) – 6\(\sqrt{63}\)
Answer:
The given expression is:
4\(\sqrt{7}\) – 6\(\sqrt{63}\)
So,
4\(\sqrt{7}\) – 6\(\sqrt{63}\)
= 4\(\sqrt{7}\) – 6\(\sqrt{7 × 9}\)
= 4\(\sqrt{7}\) – 6 (3)\(\sqrt{7}\)
= 4\(\sqrt{7}\) – 18\(\sqrt{7}\)
= \(\sqrt{7}\) ( 4 – 18)
= \(\sqrt{7}\) (-14)
= -14\(\sqrt{7}\)
Hence, from the above,
We can conclude that
4\(\sqrt{7}\) – 6\(\sqrt{63}\) = -14\(\sqrt{7}\)

Question 25.
4\(\sqrt [3]{ 5x }\) – 11\(\sqrt [3]{ 5x }\)
Answer:
The given expression is:
4\(\sqrt [3]{ 5x }\) – 11\(\sqrt [3]{ 5x }\)
So,
4\(\sqrt [3]{ 5x }\) – 11\(\sqrt [3]{ 5x }\)
= \(\sqrt [3]{ 5x }\) (4 – 11)
= \(\sqrt [3]{ 5x }\) (-7)
= -7\(\sqrt [3]{ 5x }\)
Hence, from the above,
We can conclude that
4\(\sqrt [3]{ 5x }\) – 11\(\sqrt [3]{ 5x }\) = -7\(\sqrt [3]{ 5x }\)

Question 26.
\(\sqrt{3}\)(8\(\sqrt{2}\) + 7\(\sqrt{32}\))
Answer:
The given expression is:
\(\sqrt{3}\)(8\(\sqrt{2}\) + 7\(\sqrt{32}\))
So,
\(\sqrt{3}\)(8\(\sqrt{2}\) + 7\(\sqrt{32}\))
= \(\sqrt{3}\)(8\(\sqrt{2}\) + 7\(\sqrt{16 × 2}\))
= \(\sqrt{3}\)(8\(\sqrt{2}\) + 7 (4)\(\sqrt{2}\))
= \(\sqrt{3}\)(8\(\sqrt{2}\) + 28\(\sqrt{2}\))
= \(\sqrt{3}\)(\(\sqrt{2}\) (8 + 28))
= \(\sqrt{3}\)(\(\sqrt{2}\) (36))
= 36(\(\sqrt{3}\))(\(\sqrt{2}\) )
= 36\(\sqrt{6}\)
Hence, from the above,
We can conclude that
\(\sqrt{3}\)(8\(\sqrt{2}\) + 7\(\sqrt{32}\)) = 36\(\sqrt{6}\)

Question 27.
(2\(\sqrt{5}\) – 4)2
Answer:
The given expression is:
(2\(\sqrt{5}\) – 4)²
We know that,
(a – b)² = a² – 2ab + b²
Now,
Compare the given expression with (a – b)²
We get
a = 2\(\sqrt{5}\) and b = 4
So,
(2\(\sqrt{5}\) – 4)²
= (2\(\sqrt{5}\))² – 2 (2\(\sqrt{5}\)) (4) + 4²
= 2² (\(\sqrt{5}\))² – 16\(\sqrt{5}\) + 16
= 4 (5) – 16\(\sqrt{5}\) + 16
= 20 – 16\(\sqrt{5}\) + 16
= 20 – 16 (1 – \(\sqrt{5}\))
Hence, from the above,
We can conclude that
(2\(\sqrt{5}\) – 4)²  = 20 – 16 (1 – \(\sqrt{5}\))

Question 28.
\(\sqrt [3]{ -4 }\)(\(\sqrt [3]{ 2 }\) – \(\sqrt [3]{ 16 }\))
Answer:
The given expression is:
\(\sqrt [3]{ -4 }\)(\(\sqrt [3]{ 2 }\) – \(\sqrt [3]{ 16 }\))
So,
\(\sqrt [3]{ -4 }\)(\(\sqrt [3]{ 2 }\) – \(\sqrt [3]{ 16 }\))
= \(\sqrt [3]{-4 × 2}\) – \(\sqrt [3]{-4 × 16}\)
= \(\sqrt [3]{-8}\) – \(\sqrt [3]{-64}\)
= -2 – (-4)
= -2 + 4
= 2
Hence, from the above,
We can conclude that
\(\sqrt [3]{ -4 }\)(\(\sqrt [3]{ 2 }\) – \(\sqrt [3]{ 16 }\)) = 2

Properties of Radicals

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
The process of eliminating a radical from the denominator of a radical expression is called _______________.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 1

Question 2.
VOCABULARY
What is the conjugate of the binomial \(\sqrt{6}\) + 4?
Answer:
The given expression is:
\(\sqrt{6}\) + 4
We know that,
The sum and difference of two simple quadratic surds are said to be conjugate surds to each other
Hence,
The binomial conjugate of \(\sqrt{6}\) + 4 is: \(\sqrt{6}\) – 4

Question 3.
WRITING
Are the expressions \(\frac{1}{3} \sqrt{2 x}\) and \(\sqrt{\frac{2 x}{9}}\) equivalent? Explain your reasoning.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 3

Question 4.
WHICH ONE DOESN’T BELONG?
Which expression does not belong with the other three? Explain your reasoning.
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 2
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 5–12, determine whether the expression is in simplest form. If the expression is not in simplest form, explain why.
Question 5.
\(\sqrt{19}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 5

Question 6.
\(\sqrt{\frac{1}{7}}\)
Answer:
The given expression is:
\(\sqrt{\frac{1}{7}}\)
Hence, from the above,
We can conclude that \(\sqrt{\frac{1}{7}}\) is not in its simplest form

Question 7.
\(\sqrt{48}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 7

Question 8.
\(\sqrt{34}\)
Answer:
The given expression is:
\(\sqrt{34}\)
Hence from the above,
We can conclude that \(\sqrt{34}\) is in its simplest form because the radicand has not any perfect square number

Question 9.
\(\frac{5}{\sqrt{2}}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 9

Question 10.
\(\frac{3 \sqrt{10}}{4}\)
Answer:
The given expression is:
\(\frac{3 \sqrt{10}}{4}\)
Hence, from the above,
We can conclude that \(\frac{3 \sqrt{10}}{4}\) is in its simplest form

Question 11.
\(\frac{1}{2+\sqrt[3]{2}}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 11

Question 12.
\(6-\sqrt[3]{54}\)
Answer:
The given expression is:
\(6-\sqrt[3]{54}\)
Hence from the above,
We can conclude that \(6-\sqrt[3]{54}\) is not in its simplest form because the radicand contains the perfect cube root

In Exercises 13–20, simplify the expression.
Question 13.
\(\sqrt{20}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 13

Question 14.
\(\sqrt{32}\)
Answer:
The given expression is:
\(\sqrt{32}\)
So,
\(\sqrt{32}\)
= \(\sqrt{2 × 16}\)
= \(\sqrt{2}\) × \(\sqrt{16}\)
= \(\sqrt{2}\) × 4
= 4\(\sqrt{2}\)
Hence, from the above,
We can conclude that
\(\sqrt{32}\) = 4\(\sqrt{2}\)

Question 15.
\(\sqrt{128}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 15

Question 16.
–\(\sqrt{72}\)
Answer:
The given expression is:
–\(\sqrt{72}\)
So,
–\(\sqrt{72}\)
= –\(\sqrt{36 × 2}\)
= –\(\sqrt{36}\) × \(\sqrt{2}\)
= -6 × \(\sqrt{2}\)
= -6\(\sqrt{2}\)
Hence, from the above,
We can conclude that
–\(\sqrt{72}\) = -6\(\sqrt{2}\)

Question 17.
\(\sqrt{125b}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 17

Question 18.
\(\sqrt{4 x^{2}}\)
Answer:
The given expression is:
\(\sqrt{4 x^{2}}\)
So,
\(\sqrt{4 x^{2}}\)
= \(\sqrt{(2 x)^{2}}\)
= 2x
Hence, from the above,
We can conclude that
\(\sqrt{4 x^{2}}\) = 2x

Question 19.
\(-\sqrt{81 m^{3}}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 19

Question 20.
\(\sqrt{48 n^{5}}\)
Answer:
The given expression is:
\(\sqrt{48 n^{5}}\)
So,
\(\sqrt{48 n^{5}}\)
= \(\sqrt{16 n^{4} × 3 n}\)
= \(\sqrt{16 n^{4}}\) × \(\sqrt{3 n}\)
= 4 n² × \(\sqrt{3 n}\)
= 4n²\(\sqrt{3 n}\)
Hence, from the above,
We can conclude that
\(\sqrt{48 n^{5}}\) = 4n²\(\sqrt{3 n}\)

In Exercises 21–28, simplify the expression.
Question 21.
\(\sqrt{\frac{4}{49}}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 21

Question 22.
\(-\sqrt{\frac{7}{81}}\)
Answer:
The given expression is:
\(-\sqrt{\frac{7}{81}}\)
We know that,
\(\sqrt{\frac{a}{b}}\) = \(\frac{\sqrt{a}}{\sqrt{b}}\)
So,
\(-\sqrt{\frac{7}{81}}\)
= –\(\frac{\sqrt{7}}{\sqrt{81}}\)
= –\(\frac{\sqrt{7}}{9}\)
Hence, from the above,
We can conclude that
\(-\sqrt{\frac{7}{81}}\) = –\(\frac{\sqrt{7}}{9}\)

Question 23.
\(-\sqrt{\frac{23}{64}}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 23

Question 24.
\(\sqrt{\frac{65}{121}}\)
Answer:
The given expression is:
\(\sqrt{\frac{65}{121}}\)
We know that,
\(\sqrt{\frac{a}{b}}\) = \(\frac{\sqrt{a}}{\sqrt{b}}\)
So,
\(\sqrt{\frac{65}{121}}\)
= \(\frac{\sqrt{65}}{\sqrt{121}}\)
= \(\frac{\sqrt{65}}{11}\)
Hence, from the above,
We can conclude that
\(\sqrt{\frac{65}{121}}\) = \(\frac{\sqrt{65}}{11}\)

Question 25.
\(\sqrt{\frac{a^{3}}{49}}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 25

Question 26.
\(\sqrt{\frac{144}{k^{2}}}\)
Answer:
The given expression is:
\(\sqrt{\frac{144}{k^{2}}}\)
We know that,
\(\sqrt{\frac{a}{b}}\) = \(\frac{\sqrt{a}}{\sqrt{b}}\)
So,
\(\sqrt{\frac{144}{k^{2}}}\)
= \(\frac{\sqrt{144}}{\sqrt{k^{2}}}\)
= \(\frac{12}{k}\)
Hence, from the above,
We can conclude that
\(\sqrt{\frac{144}{k^{2}}}\) = \(\frac{12}{k}\)

Question 27.
\(\sqrt{\frac{100}{4x^{2}}}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 27

Question 28.
\(\sqrt{\frac{25 v^{2}}{36}}\)
Answer:
The given expression is:
\(\sqrt{\frac{25v^{2}}{36}}\)
We know that,
\(\sqrt{\frac{a}{b}}\) = \(\frac{\sqrt{a}}{\sqrt{b}}\)
So,
\(\sqrt{\frac{25v^{2}}{36}}\)
= \(\frac{\sqrt{25v^{2}}}{\sqrt{36}}\)
= \(\frac{5v}{6}\)
Hence, from the above,
We can conclude that
\(\sqrt{\frac{25v^{2}}{36}}\) = \(\frac{5v}{6}\)

In Exercises 29–36, simplify the expression.
Question 29.
\(\sqrt [3]{ 16 }\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 29

Question 30.
\(\sqrt [3]{ -108 }\)
Answer:
The given expression is:
\(\sqrt [3]{ -108 }\)
So,
\(\sqrt [3]{ -108 }\)
= \(\sqrt [3]{ -27 × 4 }\)
= \(\sqrt [3]{ -27 }\) × \(\sqrt [3]{ 4 }\)
= -3 × \(\sqrt [3]{ 4 }\)
= -3\(\sqrt [3]{ 4 }\)
Hence, fro the above,
We can conclude that
\(\sqrt [3]{ -108 }\) = -3\(\sqrt [3]{ 4 }\)

Question 31.
\(\sqrt[3]{-64 x^{5}}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 31

Question 32.
–\(\sqrt[3]{343 n^{2}}\)
Answer:
The given expression is:
–\(\sqrt[3]{343 n^{2}}\)
So,
–\(\sqrt[3]{343 n^{2}}\)
= –\(\sqrt[3]{343 }\) × \(\sqrt[3]{ n^{2}}\)
= -7 × \(\sqrt[3]{ n^{2}}\)
= -7\(\sqrt[3]{ n^{2}}\)
Hence, from the above,
We can conclude that
–\(\sqrt[3]{343 n^{2}}\) = -7\(\sqrt[3]{ n^{2}}\)

Question 33.
\(\sqrt[3]{\frac{6 c}{-125}}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 33

Question 34.
\(\sqrt[3]{\frac{8 h^{4}}{27}}\)
Answer:
The given expression is:
\(\sqrt[3]{\frac{8 h^{4}}{27}}\)
We know that,
\(\sqrt[3]{\frac{a}{b}}\) = \(\frac{\sqrt[3]{a}}{\sqrt[3]{b}}\)
So,
\(\sqrt[3]{\frac{8 h^{4}}{27}}\)
= \(\frac{\sqrt[3]{8 h^{4}}}{\sqrt[3]{27}}\)
= \(\frac{\sqrt[3]{8 h^{3} × h}}{\sqrt[3]{27}}\)
= \(\frac{2\sqrt[3]{ h}}{3}\)
Hence, from the above,
We can conclude that
\(\sqrt[3]{\frac{8 h^{4}}{27}}\) = \(\frac{2\sqrt[3]{ h}}{3}\)

Question 35.
\(-\sqrt[3]{\frac{81 y^{2}}{1000 x^{3}}}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 35

Question 36.
\(\sqrt[3]{\frac{21}{-64 a^{3} b^{6}}}\)
Answer:
The given expression is:
\(\sqrt[3]{\frac{21}{-64 a^{3} b^{6}}}\)
We know that,
\(\sqrt[3]{\frac{a}{b}}\) = \(\frac{\sqrt[3]{a}}{\sqrt[3]{b}}\)
So,
\(\sqrt[3]{\frac{21}{-64 a^{3} b^{6}}}\)
= \(\frac{\sqrt[3]{21}}{\sqrt[3]{-64 a^{3} b^{6}}}\)
= \(\frac{\sqrt[3]{21}}{\sqrt[3]{-64 a^{3} b^{3} b^{3}}}\)
= \(\frac{\sqrt[3]{ 21}}{-4ab²}\)
Hence, from the above,
We can conclude that
\(\sqrt[3]{\frac{21}{-64 a^{3} b^{6}}}\) = \(\frac{\sqrt[3]{ 21}}{-4ab²}\)

ERROR ANALYSIS In Exercises 37 and 38, describe and correct the error in simplifying the expression.
Question 37.
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 3
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 37

Question 38.
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 4
Answer:
The given expression is:
\(\sqrt[3]{\frac{125 y^{3}}{125}}\)
We know that,
\(\sqrt[3]{\frac{a}{b}}\) = \(\frac{\sqrt[3]{a}}{\sqrt[3]{b}}\)
So,
\(\sqrt[3]{\frac{128 y^{3}}{125}}\)
= \(\frac{\sqrt[3]{128 y^{3}}}{\sqrt[3]{125}}\)
= \(\frac{\sqrt[3]{64 y^{3} × 2}}{\sqrt[3]{125}}\)
= \(\frac{4y\sqrt[3]{ 2}}{5}\)
Hence, from the above,
We can conclude that
\(\sqrt[3]{\frac{128 y^{3}}{125}}\) = \(\frac{4y\sqrt[3]{ 2}}{5}\)

In Exercises 39–44, write a factor that you can use to rationalize the denominator of the expression.
Question 39.
\(\frac{4}{\sqrt{6}}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 39

Question 40.
\(\frac{1}{\sqrt{13 z}}\)
Answer:
The given expression is:
\(\frac{1}{\sqrt{13 z}}\)
To rationalize the given expression,
We have to multiply the given expression by a factor \(\frac{\sqrt{13 z}}{\sqrt{13 z}}\)
Hence, from the above,
We can conclude that
The rationalizing factor of the given expression is: \(\frac{\sqrt{13 z}}{\sqrt{13 z}}\)

Question 41.
\(\frac{2}{\sqrt[3]{x^{2}}}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 41

Question 42.
\(\frac{3 m}{\sqrt[3]{4}}\)
Answer:
The given expression is:
\(\frac{3 m}{\sqrt[3]{4}}\)
To rationalize the given expression,
We have to multiply the given expression by a factor \(\frac{\sqrt[3]{2}}{\sqrt[3]{2}}\)
Hence, from the above,
We can conclude that
The rationalizing factor of the given expression is: \(\frac{\sqrt[3]{2}}{\sqrt[3]{2}}\)

Question 43.
\(\frac{\sqrt{2}}{\sqrt{5}-8}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 43

Question 44.
\(\frac{5}{\sqrt{3}+\sqrt{7}}\)
Answer:
The given expression is:
\(\frac{5}{\sqrt{3}+\sqrt{7}}\)
To rationalize the given expression,
We have to multiply the given expression by a factor \(\frac{\sqrt{3} – \sqrt{7}}{\sqrt{3} – \sqrt{7}}\)
Hence, from the above,
We can conclude that
The rationalizing factor of the given expression is: \(\frac{\sqrt{3} – \sqrt{7}}{\sqrt{3} – \sqrt{7}}\)

In Exercises 45–54, simplify the expression.
Question 45.
\(\frac{2}{\sqrt{2}}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 45

Question 46.
\(\frac{4}{\sqrt{3}}\)
Answer:
The given expression is:
\(\frac{4}{\sqrt{3}}\)
We have to rationalize the denominator to simplify the given expression
So,
To rationalize the denominator, multiply and divide the given fraction with \(\sqrt{3}\)
So,
\(\frac{4}{\sqrt{3}}\)
= \(\frac{4}{\sqrt{3}}\) × (\(\sqrt{3}\) / \(\sqrt{3}\))
= 4\(\sqrt{3}\) / (\(\sqrt{3}\) × \(\sqrt{3}\))
= 4\(\sqrt{3}\) / 3
Hence, from the above,
We can conclude that
\(\frac{4}{\sqrt{3}}\) = 4\(\sqrt{3}\) / 3

Question 47.
\(\frac{\sqrt{5}}{\sqrt{48}}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 47

Question 48.
\(\sqrt{\frac{4}{52}}\)
Answer:
The given expression is:
\(\sqrt{\frac{4}{52}}\)
We know that,
\(\sqrt{\frac{a}{b}}\) = \(\frac{\sqrt{a}}{\sqrt{b}}\)
So,
\(\sqrt{\frac{4}{52}}\)
= \(\frac{\sqrt{4}}{\sqrt{52}}\)
= \(\frac{\sqrt{4}}{\sqrt{4 × 13}}\)
= \(\frac{2}{2\sqrt{13}}\)
Hence, from the above,
We can conclude that
\(\sqrt{\frac{4}{52}}\) = \(\frac{2}{2\sqrt{13}}\)

Question 49.
\(\frac{3}{\sqrt{a}}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 49

Question 50.
\(\frac{1}{\sqrt{2 x}}\)
Answer:
The given expression is:
\(\frac{1}{\sqrt{2 x}}\)
We have to rationalize the denominator to simplify the given expression
So,
To rationalize the denominator, multiply and divide the given fraction with \(\sqrt{2x}\)
So,
\(\frac{1}{\sqrt{2 x}}\)
= \(\frac{1}{\sqrt{2 x}}\) × (\(\sqrt{2 x}\) / \(\sqrt{2 x}\))
= \(\sqrt{2 x}\) / (\(\sqrt{2 x}\) × \(\sqrt{2 x}\))
= \(\sqrt{2 x}\) / 2x
Hence, from the above,
We can conclude that
\(\frac{1}{\sqrt{2 x}}\) = \(\sqrt{2 x}\) / 2x

Question 51.
\(\sqrt{\frac{3 d^{2}}{5}}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 51

Question 52.
\(\frac{\sqrt{8}}{\sqrt{3 n^{3}}}\)
Answer:
The given expression is:
\(\frac{\sqrt{8}}{\sqrt{3 n^{3}}}\)
We know that,
\(\sqrt{\frac{a}{b}}\) = \(\frac{\sqrt{a}}{\sqrt{b}}\)
So,
\(\frac{\sqrt{8}}{\sqrt{3 n^{3}}}\)
= \(\frac{\sqrt{4 × 2}}{\sqrt{3 n^{2} × n}}\)
= \(\frac{2\sqrt{2}}{n\sqrt{3 n}}\)
Hence, from the above,
We can conclude that
\(\sqrt{\frac{8}{3 n^{3}}}\) = \(\frac{2\sqrt{2}}{n\sqrt{3 n}}\)

Question 53.
\(\frac{4}{\sqrt[3]{25}}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 53

Question 54.
\(\sqrt[3]{\frac{1}{108 y^{2}}}\)
Answer:
The given expression is:
\(\sqrt[3]{\frac{1}{108 y^{2}}}\)
We know that,
\(\sqrt[3]{\frac{a}{b}}\) = \(\frac{\sqrt[3]{a}}{\sqrt[3]{b}}\)
So,
\(\sqrt[3]{\frac{1}{108 y^{2}}}\)
= \(\frac{\sqrt[3]{1}}{\sqrt[3]{27 × 4 y^{2}}}\)
= \(\frac{2\sqrt[3]{1}}{3\sqrt[3]{4 y^{2}}}\)
Hence, from the above,
We can conclude that
\(\sqrt[3]{\frac{1}{108 y^{2}}}\) = \(\frac{2\sqrt[3]{1}}{3\sqrt[3]{4 y^{2}}}\)

In Exercises 55–60, simplify the expression.
Question 55.
\(\frac{1}{\sqrt{7}+1}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 55

Question 56.
\(\frac{2}{5-\sqrt{3}}\)
Answer:
The given expression is:
\(\frac{2}{5 – \sqrt{3}}\)
We have to rationalize the denominator to simplify the given expression
So,
To rationalize the denominator, multiply and divide the given fraction with \(5 + \sqrt{3}\)
So,
\(\frac{2}{5 – \sqrt{3}}\)
= \(\frac{2}{5 – \sqrt{3}}\) × (\(5 + \sqrt{3}\) / \(5 + \sqrt{3}\))
= 2\((5 + \sqrt{3})\) / (\(5 + \sqrt{3}\) × \(5 – \sqrt{3}\))
= 2\((5 + \sqrt{3})\) / 22
= \((5 + \sqrt{3})\) / 11
Hence, from the above,
We can conclude that
\(\frac{2}{5 – \sqrt{3}}\) = \((5 + \sqrt{3})\) / 11

Question 57.
\(\frac{\sqrt{10}}{7-\sqrt{2}}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 57

Question 58.
\(\frac{\sqrt{5}}{6+\sqrt{5}}\)
Answer:
The given expression is:
\(\frac{\sqrt{5}}{6 +\sqrt{5}}\)
We have to rationalize the denominator to simplify the given expression
So,
To rationalize the denominator, multiply and divide the given fraction with \(6 – \sqrt{5} \)
So,
\(\frac{\sqrt{5}}{6 + \sqrt{5}}\)
= \(\frac{\sqrt{5}}{6 + \sqrt{5}}\) × (\( 6 – \sqrt{5} \) / \(6 – \sqrt{5}\))
= \(\frac{\sqrt{5}}{6 – \sqrt{5}}\) / (\(6 +  \sqrt{5} \) × \( 6 – \sqrt{5} \))
= \(\frac{\sqrt{5}}{6 + \sqrt{5}}\) / 31
Hence, from the above,
We can conclude that
\(\frac{\sqrt{5}}{6 + \sqrt{5}}\) = \(\frac{\sqrt{5}}{6 + \sqrt{5}}\) / 31

Question 59.
\(\frac{3}{\sqrt{5}-\sqrt{2}}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 59

Question 60.
\(\frac{\sqrt{3}}{\sqrt{7}+\sqrt{3}}\)
Answer:
The given expression is:
\(\frac{\sqrt{3}}{\sqrt{7}+\sqrt{3}}\)
We have to rationalize the denominator to simplify the given expression
So,
To rationalize the denominator, multiply and divide the given fraction with \(\sqrt{7} -\sqrt{3} \)
So,
\(\frac{\sqrt{3}}{\sqrt{7}+\sqrt{3}}\)
= \(\frac{\sqrt{3}}{\sqrt{7}+\sqrt{3}}\) × (\(\sqrt{7} -\sqrt{3} \) / \(\sqrt{7} -\sqrt{3} \))
= \(\frac{\sqrt{3}}{\sqrt{7}-\sqrt{3}}\) / (\(\sqrt{7} +\sqrt{3} \) × \(\sqrt{7} -\sqrt{3} \))
= \(\frac{\sqrt{3}}{\sqrt{7}+\sqrt{3}}\) / 4
Hence, from the above,
We can conclude that
\(\frac{\sqrt{3}}{\sqrt{7}+\sqrt{3}}\) = \(\frac{\sqrt{3}}{\sqrt{7}+\sqrt{3}}\) / 4

Question 61.
MODELING WITH MATHEMATICS
The time t (in seconds) it takes an object to hit the ground is given by t = \(\sqrt{\frac{h}{16}}\), where his the height (in feet) from which the object was dropped.
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 5
a. How long does it take an earring to hit the ground when it falls from the roof of the building?
b. How much sooner does the earring hit the ground when it is dropped from two stories (22 feet) below the roof?
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 61

Question 62.
MODELING WITH MATHEMATICS
The orbital period of a planet is the time it takes the planet to travel around the Sun. You can find the orbital period P (in Earth years) using the formula P = \(\sqrt{d^{3}}\), where d is the average distance (in astronomical units, abbreviated AU) of the planet from the Sun.
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 6
a. Simplify the formula.
Answer:
It is given that
You can find the orbital period P (in Earth years) using the formula
P = \(\sqrt{d^{3}}\)
where d is the average distance (in astronomical units, abbreviated AU)
Now,
The given formula is:
P = \(\sqrt{d^{3}}\)
So,
P = \(\sqrt{d^{3}}\)
P = \(\sqrt{d^{2} × d}\)
P = d\(\sqrt{d}\)
Hence, from the above,
We can conclude that
The simplified formula for the orbital period(P) is:
P = d\(\sqrt{d}\)

b. What is Jupiter’s orbital period?
Answer:
From part (a),
The simplified formula for the Orbital period (P) is:
P = d\(\sqrt{d}\)
It is given that
The distance of Jupiter from the sun is: 5.2 AU
So,
The Orbital Period of Jupiter (P) is:
P = \(\sqrt{5.2^{3}}\)
P = 11.85 AU
Hence, from the above,
We can conclude that the Orbital Period of Jupiter (P) is: 11.85 AU

Question 63.
MODELING WITH MATHEMATICS
The electric current I (in amperes) an appliance uses is given by the formula I = \(\sqrt{\frac{P}{R}}\), where P is the power (in watts) and R is the resistance (in ohms). Find the current an appliance uses when the power is 147 watts and the resistance is 5 ohms.
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 7
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 63

Question 64.
MODELING WITH MATHEMATICS
You can find the average annual interest rate r (in decimal form) of a savings account using the formula r = \(\sqrt{\frac{V_{2}}{V_{0}}}\) – 1, where V0 is the initial investment and V2 is the balance of the account after 2 years. Use the formula to compare the savings accounts. In which account would you invest money? Explain.
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 8
Answer:
It is given that
You can find the average annual interest rate r (in decimal form) of a savings account using the formula
r = \(\sqrt{\frac{V_{2}}{V_{0}}}\) – 1
where,
V0 is the initial investment
V2 is the balance of the account after 2 years
We know that,
We would invest money in the account where there is more average annual interest when compared to other accounts
So,
Now,
For \(\frac{V_{2}}{V_{0}}\) = \(\frac{293}{275}\),
r = \(\sqrt{\frac{293}{275}}\) – 1
r = 3.19%
For \(\frac{V_{2}}{V_{0}}\) = \(\frac{382}{361}\),
r = \(\sqrt{\frac{382}{361}}\) – 1
r = 2.85%
For \(\frac{V_{2}}{V_{0}}\) = \(\frac{214}{199}\),
r = \(\sqrt{\frac{214}{199}}\) – 1
r = 3.68%
For \(\frac{V_{2}}{V_{0}}\) = \(\frac{272}{254}\),
r = \(\sqrt{\frac{272}{254}}\) – 1
r = 3.44%
For \(\frac{V_{2}}{V_{0}}\) = \(\frac{406}{386}\),
r = \(\sqrt{\frac{406}{386}}\) – 1
r = 2.51%
Hence, from the above,
We can conclude that we would invest money in Account 3 since the average annual interest rate is high for this account when compared to other accounts

In Exercises 65–68, evaluate the function for the given value of x. Write your answer in simplest form and in decimal form rounded to the nearest hundredth.
Question 65.
h(x) = \(\sqrt{5x}\); x = 10
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 65

Question 66.
g(x) = \(\sqrt{3x}\); x = 60
Answer:
The given function is:
g(x) = \(\sqrt{3x}\) with x = 60
So,
g(60) = \(\sqrt{3 × 60}\)
g(60) = \(\sqrt{9 × 20}\)
g(60) = \(\sqrt{9 × 5 × 4}\)
g(60) = \(\sqrt{36 × 5}\)
g(60) = 6\(\sqrt{5}\)
g(60) = 13.41
Hence, from the above,
e can conclude that the value of the given function is: 13.41

Question 67.
r(x) = \(\sqrt{\frac{3 x}{3 x^{2}+6}}\); x = 4
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 67

Question 68.
p(x) = \(\sqrt{\frac{x-1}{5 x}}\); x = 8
Answer:
The given function is:
p(x) = \(\sqrt{\frac{x-1}{5 x}}\) with x = 8
So,
p(8) = \(\sqrt{\frac{8-1}{5 × 8}}\)
p(8) = \(\sqrt{\frac{7}{40}}\)
p(8) = 0.41
Hence, from the above,
We can conclude that the value of the given function is: 0.41

In Exercises 69–72, evaluate the expression when a = −2, b = 8, and c = \(\frac{1}{2}\). Write your answer in simplest form and in decimal form rounded to the nearest hundredth.
Question 69.
\(\sqrt{a^{2}+b c}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 69

Question 70.
\(-\sqrt{4 c-6 a b}\)
Answer:
The given expression is:
\(-\sqrt{4 c-6 a b}\)
So,
\(-\sqrt{4 c-6 a b}\)
= \(-\sqrt{4 × (1/2) – 6 (-2) (8)}\)
= \(-\sqrt{2- 6 (-16)}\)
= \(-\sqrt{2 + 96}\)
= \(-\sqrt{98}\)
= -9.89
Hence, from the above,
We can conclude that the value of the given expression is: -9.89

Question 71.
\(-\sqrt{2 a^{2}+b^{2}}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 71

Question 72.
\(\sqrt{b^{2}-4 a c}\)
Answer:
The given expression is:
\(\sqrt{b^{2}-4 a c}\)
So,
\(\sqrt{b^{2}-4 a c}\)
= \(\sqrt{8^{2}-4 (-2) × (1/2)}\)
= \(\sqrt{64 + 4}\)
= \(\sqrt{68}\)
= 8.24
Hence, from the above,
We can conclude that the value of the given expression is: 8.24

Question 73.
MODELING WITH MATHEMATICS
The text in the book shown forms a golden rectangle. What is the width w of the text?
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 9
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 73

Question 74.
MODELING WITH MATHEMATICS
The flag of Togo is approximately the shape of a golden rectangle. What is the width w of the flag?
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 10
Answer:
It is given that
The flag of Togo is approximately the shape of a golden rectangle
It is also given that
The length of the golden rectangle is 42 in.
The width of the given rectangle is w inch
Now,
We know that,
The ratio of the length and the width in a golden rectangle = \(\frac{1+\sqrt{5}}{2}\)
So,
\(\frac{1+\sqrt{5}}{2}\) = \(\frac{42}{w}\)
w (\(1 + \sqrt{5}\)) = 42 × 2
h (\(1 + \sqrt{5}\)) = 84
h = \(\frac{84}{1 + \sqrt{5}}\)
We have to rationalize the denominator to simplify the given expression
So,
To rationalize the denominator, multiply and divide the given fraction with \(1 – \sqrt{5}\)
So,
\(\frac{84}{1 + \sqrt{5}}\)
= \(\frac{84}{1 + \sqrt{5}}\) × (\(1 – \sqrt{5}\) / \(1 – \sqrt{5}\))
= 84\((1 – \sqrt{5})\) / (\(1 + \sqrt{5}\) × \(1 – \sqrt{5}\))
= 84\((1 – \sqrt{5})\) / -4
= – 84\((1 – \sqrt{5})\) / 4
= 25.95 in
Hence, from the above,
We can conclude that the width of the golden rectangle is 25.95 in

In Exercises 75–82, simplify the expression.
Question 75.
\(\sqrt{2}\) – 2\(\sqrt{2}\) + 6\(\sqrt{2}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 75

Question 76.
\(\sqrt{5}\) – 5\(\sqrt{13}\) – 8\(\sqrt{5}\)
Answer:
The given expression is:
\(\sqrt{5}\) – 5\(\sqrt{13}\) – 8\(\sqrt{5}\)
So,
\(\sqrt{5}\) – 5\(\sqrt{13}\) – 8\(\sqrt{5}\)
= \(\sqrt{5}\) (1 – 8) – 5\(\sqrt{13}\)
= \(\sqrt{5}\) (-7) – 5\(\sqrt{13}\)
= -7\(\sqrt{5}\)  – 5\(\sqrt{13}\)
Hence, from the above,
We can conclude that
\(\sqrt{5}\) – 5\(\sqrt{13}\) – 8\(\sqrt{5}\) = -7\(\sqrt{5}\)  – 5\(\sqrt{13}\)

Question 77.
2\(\sqrt{6}\) – 5\(\sqrt{54}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 77

Question 78.
9\(\sqrt{32}\) + \(\sqrt{2}\)
Answer:
The given expression is:
9\(\sqrt{32}\) + \(\sqrt{2}\)
So,
9\(\sqrt{32}\) + \(\sqrt{2}\)
= 9\(\sqrt{16 × 2}\) + \(\sqrt{2}\)
= 9 (4)\(\sqrt{2}\) + \(\sqrt{2}\)
= 36\(\sqrt{2}\) + \(\sqrt{2}\)
= \(\sqrt{2}\) (6 + 1)
= \(\sqrt{2}\) (7)
= 7\(\sqrt{2}\)
Hence, from the above,
We can conclude that
9\(\sqrt{32}\) + \(\sqrt{2}\) = 7\(\sqrt{2}\)

Question 79.
\(\sqrt{12}\) + 6\(\sqrt{3}\) + 2\(\sqrt{6}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 79

Question 80.
3\(\sqrt{7}\) – 5\(\sqrt{14}\) + 2\(\sqrt{28}\)
Answer:
The given expression is:
3\(\sqrt{7}\) – 5\(\sqrt{14}\) + 2\(\sqrt{28}\)
So,
3\(\sqrt{7}\) – 5\(\sqrt{14}\) + 2\(\sqrt{28}\)
= 3\(\sqrt{7}\) – 5\(\sqrt{14}\) + 2\(\sqrt{7 × 4}\)
= 3\(\sqrt{7}\) – 5\(\sqrt{14}\) + 2 (2)\(\sqrt{7}\)
= 3\(\sqrt{7}\) – 5\(\sqrt{14}\) + 4\(\sqrt{7}\)
= \(\sqrt{7}\) (3 + 4) – 5\(\sqrt{14}\)
= \(\sqrt{7}\) (7) – 5\(\sqrt{14}\)
= 7\(\sqrt{7}\) – 5\(\sqrt{14}\)
Hence, from the above,
We can conclude that
3\(\sqrt{7}\) – 5\(\sqrt{14}\) + 2\(\sqrt{28}\) = 7\(\sqrt{7}\) – 5\(\sqrt{14}\)

Question 81.
\(\sqrt[3]{-81}\) + 4\(\sqrt[3]{3}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 81

Question 82.
6\(\sqrt[3]{128 t}\) – 2\(\sqrt[3]{2 t}\)
Answer:
The given expression is:
6\(\sqrt[3]{128 t}\) – 2\(\sqrt[3]{2 t}\)
So,
6\(\sqrt[3]{128 t}\) – 2\(\sqrt[3]{2 t}\)
= 6\(\sqrt[3]{64 × 2 t}\) – 2\(\sqrt[3]{2 t}\)
= 6 (4)\(\sqrt[3]{2 t}\) – 2\(\sqrt[3]{2 t}\)
= 24\(\sqrt[3]{2 t}\) – 2\(\sqrt[3]{2 t}\)
= \(\sqrt[3]{2 t}\) (24 – 2)
= \(\sqrt[3]{2 t}\) (22)
= 22\(\sqrt[3]{2 t}\)
Hence, from the above,
We can conclude that
6\(\sqrt[3]{128 t}\) – 2\(\sqrt[3]{2 t}\) = 22\(\sqrt[3]{2 t}\)

In Exercises 83–90, simplify the expression.
Question 83.
\(\sqrt{2}\)(\(\sqrt{45}\) + \(\sqrt{5}\))
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 83

Question 84.
\(\sqrt{3}\)(\(\sqrt{72}\) – 3\(\sqrt{2}\))
Answer:
The given expression is:
\(\sqrt{3}\)(\(\sqrt{72}\) – 3\(\sqrt{2}\))
So,
\(\sqrt{3}\)(\(\sqrt{72}\) – 3\(\sqrt{2}\))
= \(\sqrt{3}\)(\(\sqrt{36 × 2}\) – 3\(\sqrt{2}\))
= \(\sqrt{3}\)(6\(\sqrt{2}\) – 3\(\sqrt{2}\))
= \(\sqrt{3}\)(\(\sqrt{2}\) (6 – 3))
= \(\sqrt{3}\)(\(\sqrt{2}\) (3))
= \(\sqrt{3}\)(3\(\sqrt{2}\))
= 3 ×  \(\sqrt{3}\) × \(\sqrt{2}\)
= 3 \(\sqrt{6}\)
Hence, from the above,
We can conclude that
\(\sqrt{3}\)(\(\sqrt{72}\) – 3\(\sqrt{2}\)) = 3 \(\sqrt{6}\)

Question 85.
\(\sqrt{5}\)(2\(\sqrt{6x}\) – \(\sqrt{96x}\))
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 85

Question 86.
\(\sqrt{7y}\)(\(\sqrt{27y}\) + 5\(\sqrt{12y}\))
Answer:
The given expression is:
\(\sqrt{7y}\)(\(\sqrt{27y}\) + 5\(\sqrt{12y}\))
So,
\(\sqrt{7y}\)(\(\sqrt{27y}\) + 5\(\sqrt{12y}\))
= \(\sqrt{7y}\)(\(\sqrt{9 × 3y}\) + 5\(\sqrt{4 × 3y}\))
= \(\sqrt{7y}\)(3\(\sqrt{3y}\) + 5 (2)\(\sqrt{3y}\))
= \(\sqrt{7y}\)(3\(\sqrt{3y}\) + 10\(\sqrt{3y}\))
= \(\sqrt{7y}\)(\(\sqrt{3y}\) (3 + 10))
= \(\sqrt{7y}\)(\(\sqrt{3y}\) (13))
= 13 × \(\sqrt{7y}\) × \(\sqrt{3y}\)
= 13(\(\sqrt{21y}\)
Hence, from the above,
We can conclude that
\(\sqrt{7y}\)(\(\sqrt{27y}\) + 5\(\sqrt{12y}\)) = 13(\(\sqrt{21y})\)

Question 87.
(4\(\sqrt{2}\) – \(\sqrt{98}\))2
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 87

Question 88.
(\(\sqrt{3}\) + \(\sqrt{48}\)) (\(\sqrt{20}\) – \(\sqrt{5}\))
Answer:
The given expression is:
(\(\sqrt{3}\) + \(\sqrt{48}\)) (\(\sqrt{20}\) – \(\sqrt{5}\))
So,
(\(\sqrt{3}\) + \(\sqrt{48}\)) (\(\sqrt{20}\) – \(\sqrt{5}\))
= (\(\sqrt{3}\) + \(\sqrt{16 × 3}\)) (\(\sqrt{5 × 4}\) – \(\sqrt{5}\))
= (\(\sqrt{3}\) + 4\(\sqrt{3}\)) (2\(\sqrt{5}\) – \(\sqrt{5}\))
= (\(\sqrt{3}\) (1 + 4)) (\(\sqrt{5}\) (2 – 1))
= (\(\sqrt{3}\) (5)) (\(\sqrt{5}\) (1))
= 5\(\sqrt{3}\) × \(\sqrt{5}\)
= 5\(\sqrt{15}\)
Hence, from the above,
We can conclude that
(\(\sqrt{3}\) + \(\sqrt{48}\)) (\(\sqrt{20}\) – \(\sqrt{5}\)) = 5\(\sqrt{15}\)

Question 89.
\(\sqrt[3]{3}(\sqrt[3]{4}+\sqrt[3]{32})\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 89

Question 90.
\(\sqrt[3]{2}(\sqrt[3]{135}-4 \sqrt[3]{5})\)
Answer:
The given expression is:
\(\sqrt[3]{2}(\sqrt[3]{135}-4 \sqrt[3]{5})\)
So,
\(\sqrt[3]{2}(\sqrt[3]{135}-4 \sqrt[3]{5})\)
= \(\sqrt[3]{2}(\sqrt[3]{5 × 27}-4 \sqrt[3]{5})\)
= \(\sqrt[3]{2}(3\sqrt[3]{5}-4 \sqrt[3]{5})\)
= \(\sqrt[3]{2}(\sqrt[3]{5} (3 – 4)\)
= \(\sqrt[3]{2}(\sqrt[3]{5} (-1)\)
= –\(\sqrt[3]{2}(\sqrt[3]{5}) \)
= –\(\sqrt[3]{10}\)
Hence, from the above,
We can conclude that
\(\sqrt[3]{2}(\sqrt[3]{135}-4 \sqrt[3]{5})\) = –\(\sqrt[3]{10}\)

Question 91.
MODELING WITH MATHEMATICS
The circumference C of the art room in a mansion is approximated by the formula C ≈ \(\sqrt\frac{a^{2}+b^{2}}{2}\). Approximate the circumference of the room.
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 11
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 91

Question 92.
CRITICAL THINKING
Determine whether each expression represents a rational or an irrational number. Justify your answer.
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 12
Answer:
a.
The given expression is:
4 + \(\sqrt{6}\)
We know that,
Any fraction that we can write in the form of \(\frac{p}{q}\) is called a “Rational Number”. Otherwise, that number is an “Irrational number”
So,
We can say that the given expression can’t be expressed in the form of \(\frac{p}{q}\)
Hence, from the above,
We can conclude that 4 + \(\sqrt{6}\) is an irrational number
b.
The given expression is:
\(\frac{\sqrt{48}}{\sqrt{3}}\)
We know that,
Any fraction that we can write in the form of \(\frac{p}{q}\) is called a “Rational Number”. Otherwise, that number is an “Irrational number”
So,
We can say that the given expression can be expressed in the form of \(\frac{p}{q}\)
Hence, from the above,
We can conclude that \(\frac{\sqrt{48}}{\sqrt{3}}\) is a rational number
c.
The given expression is:
\(\frac{8}{\sqrt{12}}\)
We know that,
Any fraction that we can write in the form of \(\frac{p}{q}\) is called a “Rational Number”. Otherwise, that number is an “Irrational number”
So,
We can say that the given expression can’t be expressed in the form of \(\frac{p}{q}\)
Hence, from the above,
We can conclude that \(\frac{8}{\sqrt{12}}\) is an irrational number
d.
The given expression is:
\(\frac{\sqrt{3}}{\sqrt{7}}\)
We know that,
Any fraction that we can write in the form of \(\frac{p}{q}\) is called a “Rational Number”. Otherwise, that number is an “Irrational number”
So,
We can say that the given expression can’t be expressed in the form of \(\frac{p}{q}\)
Hence, from the above,
We can conclude that\(\frac{\sqrt{3}}{\sqrt{7}}\) is an irrational number
e.
The given expression is:
\(\frac{a}{\sqrt{10} – \sqrt{2}}\)
We know that,
Any fraction that we can write in the form of \(\frac{p}{q}\) is called a “Rational Number”. Otherwise, that number is an “Irrational number”
So,
We can say that the given expression can be expressed in the form of \(\frac{p}{q}\)
Hence, from the above,
We can conclude that\(\frac{a}{\sqrt{10} – \sqrt{2}}\) is a rational number
f.
The given expression is:
\(\frac{2 + \sqrt{5}}{2 b + \sqrt{5 b^{2}}}\)
We know that,
Any fraction that we can write in the form of \(\frac{p}{q}\) is called a “Rational Number”. Otherwise, that number is an “Irrational number”
So,
We can say that the given expression can be expressed in the form of \(\frac{p}{q}\)
Hence, from the above,
We can conclude that\(\frac{2 + \sqrt{5}}{2 b + \sqrt{5 b^{2}}}\) is a rational number

In Exercises 93–98, simplify the expression.
Question 93.
\(\sqrt[5]{\frac{13}{5 x^{5}}}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 93

Question 94.
\(\sqrt[4]{\frac{10}{81}}\)
Answer:
The given expression is:
\(\sqrt[4]{\frac{10}{81}}\)
We know that,
\(\sqrt[4]{\frac{a}{b}}\) = \(\frac{\sqrt[4]{a}}{\sqrt[4]{b}}\)
So,
\(\sqrt[4]{\frac{10}{81}}\)
= \(\frac{\sqrt[4]{10}}{\sqrt[4]{81}}\)
= \(\frac{\sqrt[4]{10}}{\sqrt[4]{3^{4}}}\)
= \(\frac{\sqrt[4]{10}}{3}\)
Hence, from the above,
We can conclude that
\(\sqrt[4]{\frac{10}{81}}\) = \(\frac{\sqrt[4]{10}}{3}\)

Question 95.
\(\sqrt[4]{256 y}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 95

Question 96.
\(\sqrt[5]{160 x^{6}}\)
Answer:
The given expression is:
\(\sqrt[5]{160 x^{6}}\)
So,
\(\sqrt[5]{160 x^{6}}\)
= \(\sqrt[5]{40 (4) x^{5} × x}\)
= \(\sqrt[5]{20 (8) x^{5} × x}\)
= \(\sqrt[5]{32 (5) x^{5} × x}\)
= \(\sqrt[5]{2^{5} × 5  x^{5} × x}\)
= 2x\(\sqrt[5]{5 x}\)
Hence, from the above,
We can conclude that
\(\sqrt[5]{160 x^{6}}\) = 2x\(\sqrt[5]{5 x}\)

Question 97.
\(6 \sqrt[4]{9}-\sqrt[5]{9}+3 \sqrt[4]{9}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 97

Question 98.
\(\sqrt[5]{2}(\sqrt[4]{7}+\sqrt[5]{16})\)
Answer:
The given expression is:
\(\sqrt[5]{2}(\sqrt[4]{7}+\sqrt[5]{16})\)
So,
\(\sqrt[5]{2}(\sqrt[4]{7}+\sqrt[5]{16})\)
= (\(\sqrt[5]{2}\) × \(\sqrt[4]{7}\)) + (\(\sqrt[5]{2}\) × \(\sqrt[5]{16}\))
= (\(\sqrt[5]{2}\) × \(\sqrt[4]{7}\)) + (\(\sqrt[5]{32}\)
= (\(\sqrt[5]{2}\) × \(\sqrt[4]{7}\)) + 2
Hence, from the above,
We can conclude that
\(\sqrt[5]{2}(\sqrt[4]{7}+\sqrt[5]{16})\) = (\(\sqrt[5]{2}\) × \(\sqrt[4]{7}\)) + 2

REASONING In Exercises 99 and 100, use the table shown.
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 13
Question 99.
Copy and complete the table by (a) finding each sum ( 2 + 2, 2 + \(\frac{1}{4}\), etc. ) and (b) finding each product ( 2 • 2, 2 • \(\frac{1}{4}\), etc. )
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 99.1
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 99.2

Question 100.
Use your answers in Exercise 99 to determine whether each statement is always, sometimes, or never true. Justify your answer.
a. The sum of a rational number and a rational number is rational.
b. The sum of a rational number and an irrational number is irrational.
c. The sum of an irrational number and an irrational number is irrational.
d. The product of a rational number and a rational number is rational.
e. The product of a nonzero rational number and an irrational number is irrational.
f. The product of an irrational number and an irrational number is irrational.
Answer:
From Exercise 99,
a. The completed table of the sum of rational numbers and irrational numbers is:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 99.1
b. The completed table of the product of rational numbers and irrational numbers is:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 99.2
Now,
The given statements are:
a. The sum of a rational number and a rational number is rational.
b. The sum of a rational number and an irrational number is irrational.
c. The sum of an irrational number and an irrational number is irrational.
d. The product of a rational number and a rational number is rational.
e. The product of a nonzero rational number and an irrational number is irrational.
f. The product of an irrational number and an irrational number is irrational.
So,
a.
The given statement is sometimes true
b.
The given statement is sometimes true
c.
The given statement is always true
d.
The given statement is always true
e.
The given statement is sometimes true
f.
The given statement is sometimes true

Question 101.
REASONING
Let m be a positive integer. For what values of m will the simplified form of the expression \(\sqrt{2^{m}}\) contain a radical? For what values will it not contain a radical? Explain.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 101

Question 102.
HOW DO YOU SEE IT?
The edge length s of a cube is an irrational number, a surface area is an irrational number, and a volume is a rational number. Give a possible value of s.
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 14
Answer:
It is given that
The length (s) of a cube is: An irrational number
The surface area of a cube is: An irrational number
The volume of a cube is: A rational number
Now,
We have to find the possible value of s i.e., the length of the side of a cube
Now,
We know that,
The surface area of a cube = 6 (Side)²
The volume of a cube = (Side)³
Now,
Let the length of the side of the cube = \(\sqrt{2}\)
So,
The surface area of a cube = 6 (√2)²
= 6 (2)
= 12
The volume of a cube = (√2)³
So,
From the above values,
We can observe that the surface area of a cube will always be rational irrespective the number is rational or irrational
Hence, from the above,
We can conclude that there is no possible value of s

Question 103.
REASONING
Let a and b be positive numbers. Explain why \(\sqrt{ab}\) lies between a and b on a number line. (Hint: Let a< b and multiply each side of a < b by a. Then let a < b and multiply each side by b.)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 103

Question 104.
MAKING AN ARGUMENT
Your friend says that you can rationalize the denominator of the expression \(\frac{2}{4+\sqrt[3]{5}}\) by multiplying the numerator and denominator by 4 – \(\sqrt[3]{5}\). Is your friend correct? Explain.
Answer:
No, your friend is not correct

Explanation:
It is given that
You can rationalize the denominator of the expression \(\frac{2}{4+\sqrt[3]{5}}\) by multiplying the numerator and denominator by 4 – \(\sqrt[3]{5}\).
We know that,
We will rationalize a radical to make that radical a rational number
A “Rational number” is a number that can be written in the form of \(\frac{p}{q}\)
So,
Now,
We will see whether we can rationalize the denominator or not in the given fraction
So,
(4 – \(\sqrt[3]{5}\)) × (4 + \(\sqrt[3]{5}\))
= 4² – (\(\sqrt[3]{5}\) × \(\sqrt[3]{5}\))
= 16 – (\(\sqrt[3]{5}\) × \(\sqrt[3]{5}\))
So,
From the above value,
We can observe that the denominator can’t be rationalized
Hence, from the above,
We can conclude that your friend is not correct

Question 105.
PROBLEM -SOLVING
The ratio of consecutive terms \(\frac{a_{n}}{a_{n}-1}\) in the Fibonacci sequence gets closer and closer to the golden ratio \(\frac{1+\sqrt{5}}{2}\) as n increases. Find the term that precedes 610 in the sequence.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 105.1
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 105.2

Question 106.
THOUGHT-PROVOKING
Use the golden ratio \(\frac{1+\sqrt{5}}{2}\) and the golden ratio conjugate \(\frac{1-\sqrt{5}}{2}\) for each of the following.
a. Show that the golden ratio and golden ratio conjugate are both solutions of x2 – x – 1 = 0.
Answer:
The given equation is:
x² – x – 1 = 0
We know that,
To find the solutions to the given equation,
We have to use the below formula:
Solution1, Solution 2 = \(\frac{-b + \sqrt{b² – 4 a c}}{2 a}\), \(\frac{-b – \sqrt{b² – 4 a c}}{2 a}\)
Now,
Compare the give equation with
ax² +  x + c = 0
We get,
a = 1, b = -1, and c = -1
So,
Solution1, Solution 2 = \(\frac{1 + \sqrt{(-1)² – 4 (1) (-1)}}{2 (1)}\), \(\frac{1 – \sqrt{(-1)² – 4 (1) (-1)}}{2 (1)}\)
Solution1, Solution 2 = \(\frac{1 + \sqrt{1 + 4}}{2}\), \(\frac{1 – \sqrt{1 + 4}}{2}\)
Solution1, Solution 2 = \(\frac{1 + \sqrt{5}}{2}\), \(\frac{1 – \sqrt{5}}{2}\)
Hence, from the above,
We can conclude that the golden ratio and golden ratio conjugate are both solutions of the given equation

b. Construct a geometric diagram that has the golden ratio as the length of a part of the diagram.
Answer:
It is given that
The golden ratio is: \(\frac{1 + \sqrt{5}}{2}\)
Now,
\(\frac{1 + \sqrt{5}}{2}\) = 1.61
Hence,
The representation of the golden ratio as the length of the part of the diagram is:

Question 107.
CRITICAL THINKING
Use the special product pattern (a + b)(a2 – ab + b2) = a3 + b3 to simplify the expression \(\frac{2}{\sqrt[3]{x}+1}\). Explain your reasoning.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 107

Maintaining Mathematical Proficiency

Graph the linear equation. Identify the x-intercept.
Question 108.
y = x – 4
Answer:
The given linear equation is:
y = x – 4
Compare the given equation with
y = mx + c
Where,
m is the slope
c is the y-intercept
To find the x-intercept,put y = 0
So,
x – 4 = 0
x = 4
Hence, from the above,
We can conclude that
The graph of the linear equation is

The x-intercept of the linear equation is: 4

Question 109.
y = -2x + 6
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 109

Question 110.
y = –\(\frac{1}{3}\)x – 1
Answer:
The given linear equation is:
y = –\(\frac{1}{3}\)x – 1
Compare the given equation with
y = mx + c
Where,
m is the slope
c is the y-intercept
To find the x-intercept,put y = 0
So,
–\(\frac{1}{3}\)x – 1 = 0
–\(\frac{1}{3}\)x = 1
x = -3
Hence, from the above,
We can conclude that
The graph of the linear equation is

The x-intercept of the linear equation is: -3

Question 111.
y = \(\frac{3}{2}\)x + 6
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 111

Solve the equation. Check your solution.
Question 112.
32 = 2x
Answer:
The given expression is:
32 = 2^x
2^5 = 2^x
We know that,
If the bases are equal, then the exponents can also be equal
So,
x = 5
Hence, from the above,
We can conclude that the solution of the given equation is: x= 5

Question 113.
27x = 3x – 6
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 113

Question 114.
(\(\frac{1}{6}\))2x = 2161 – x
Answer:
The given expression is:
(\(\frac{1}{6}\))2x = 2161 – x
We know that,
\(\frac{1}{x}\) = x^-1
So,
[6^-1]^2 x = [6^3]^1 – x
6^[-2x] = 6^[3(1 – x)]
We know that,
if the bases are equal, then the exponents are equal
So,
-2x = 3 (1 – x)
-2x = 3 – 3x
-2x + 3x = 3
x = 3
Hence, from the above,
We can conclude that the solution of the given equation is: x = 3

Question 115.
625x = (\(\frac{1}{25}\))x + 2
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.1 a 115

Lesson 9.2 Solving Quadratic Equations by Graphing

Essential Question How can you use a graph to solve a quadratic equation in one variable?
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 1
Based on what you learned about the x-intercepts of a graph in Section 3.4, it follows that the x-intercept of the graph of the linear equation
y = ax + b —-> 2 variables
is the same value as the solution of
ax + b = 0 —–> 1 variable
You can use similar reasoning to solve quadratic equations.

EXPLORATION 1

Solving a Quadratic Equation by Graphing
Work with a partner.
a. Sketch the graph of y = x2 – 2x.
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 2
Answer:
The given equation is:
y = x² – 2x
Hence,
The graph of the given equation in the coordinate plane is:

b. What is the definition of an x-intercept of a graph? How many x-intercepts does this graph have? What are they?
Answer:
The x-intercept is the point where a line crosses the x-axis,
Hence,
From the graph of part (a),
We can observe that there are 2 points where a line crosses the x-axis
The 2 points that a line crosses the x-axis are: 0, 2
Hence, from the above,
We can conclude that there are two x-intercepts in this graph i.e., 0 and 2

c. What is the definition of a solution of an equation in x? How many solutions does the equation x2 – 2x = 0 have? What are they?
Answer:
A solution is an assignment of values to the unknown variables that make the equality in the equation true
So,
To find the number of solutions,
Make the equation equal to 0
So,
x² – 2x = 0
x (x – 2) = 0
x = 0 or x – 2 = 0
x = 0 or x = 0 + 2
x = 0 or x = 2
Hence, from the above,
We can conclude that there are 2 solutions for the given equation
The 2 solutions of the given equation are: 0 and 2

d. Explain how you can verify the solutions you found in part (c).
Answer:
We can verify the solutions by substituting the solutions we got in the given equation
So,
From part (c),
the solutions we got are: 0 and 2
So,
Now,
Put x = 0 in the given equation
0 – 2 (0) = 0
0 = 0
Put x = 2 in the given equation
2² – 2 (2) = 0
4 – 4 = 0
0 = 0
Hence,
In the above way,
We can verify the solutions that we found in part (c)

EXPLORATION 2

Solving Quadratic Equations by Graphing
Work with a partner. Solve each equation by graphing.
a. x2 – 4 = 0
Answer:
The given equation is:
x2 – 4 = 0
Hence,
The graph of the given equation in the coordinate plane is:

From the above graph,
The solutions that satisfy the above equation i.e., x-intercepts are: -2 and 2
Hence, from the above,
We can conclude that the solutions of the given equation are: -2 and 2
b. x2 + 3x = 0
Answer:
The given equation is:
x2 + 3x = 0
Hence,
The graph of the given equation in the coordinate plane is:

From the above graph,
The solutions that satisfy the above equation i.e., x-intercepts are: 0 and -3
Hence, from the above,
We can conclude that the solutions of the given equation are: 0 and -3

c. -x2 + 2x = 0
Answer:
The given equation is:
-x2 + 2x = 0
Hence,
The graph of the given equation in the coordinate plane is:

From the above graph,
The solutions that satisfy the above equation i.e., x-intercepts are: 0 and -2
Hence, from the above,
We can conclude that the solutions of the given equation are: 0 and -2

d. x2 – 2x + 1 = 0
Answer:
The given equation is:
x2 – 2x + 1 = 0
Hence,
The graph of the given equation in the coordinate plane is:

From the above graph,
The solution that satisfies the above equation i.e., x-intercepts is: 1
Hence, from the above,
We can conclude that the solution of the given equation is: 1

e. x2 – 3x + 5 = 0
Answer:
The given equation is:
x2 – 3x + 5 = 0
Hence,
The graph of the given equation in the coordinate plane is:

From the above graph,
The approximate solutions that satisfy the above equation i.e., x-intercepts are: 0.9 and 2.1
Hence, from the above,
We can conclude that the approximate solutions of the given equation are: 0.9 and 2.1

f. -x2 + 3x – 6 = 0
Answer:
The given equation is:
-x2 + 3x – 6 = 0
Hence,
The graph of the given equation in the coordinate plane is:

From the above graph,
The solutions that satisfy the above equation i.e., x-intercepts are: 1 and 2
Hence, from the above,
We can conclude that the solutions of the given equation are: 1 and 2

Communicate Your Answer

Question 3.
How can you use a graph to solve a quadratic equation in one variable?
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 3
Answer:
Based on what you learned about the x-intercepts of a graph in Section 3.4, it follows that the x-intercept of the graph of the linear equation
y = ax + b —-> 2 variables
is the same value as the solution of
ax + b = 0 —–> 1 variable
You can use similar reasoning to solve quadratic equations.

Question 4.
After you find a solution graphically, how can you check your result algebraically? Check your solutions for parts (a)-(d) in Exploration 2 algebraically.
Answer:
After finding the solutions graphically for the equations of Exploration 2,
We can check whether the solutions are true or not algebraically by substituting the solutions in the equations
So,
Now,
a.
The given equation is:
x² – 4 = 0
From the graph,
The solutions of the given equation are: -2 and 2
So,
(-2)² -4 = 0
4 – 4 = 0
2² – 4 = 0
4 – 4 = 0
0 = 0
b.
The given equation is:
x² + 3x = 0
From the graph,
The solutions of the given equation are: 0 and -3
So,
0² + 3 (0) = 0
0 + 0 = 0
0 = 0
(-3)² + 3 (-3) = 0
9 – 9 = 0
0 = 0
c.
The given equation is:
-x² + 2x = 0
From the graph,
The solutions of the given equation are: 0 and 2
So,
0² + 2 (0) = 0
0 + 0 = 0
0 = 0
-(2)² + 2 (2) = 0
4 – 4 = 0
0 = 0
d.
The given equation is:
x² – 2x + 1 = 0
From the graph,
The solution of the given equation are: 1
So,
1² – 2 (1) + 1 = 0
1 – 2 + 1 = 0
2 – 2 = 0
0 = 0

Question 5.
How can you determine graphically that a quadratic equation has no solution?
Answer:
Looking at the graph of a quadratic equation, if the parabola does not cross or intersect the x-axis, then the equation has no real solution. And no real solution does not mean that there is no solution, but that the solutions are not real numbers.

Monitoring Progress

Solve the equation by graphing. Check your solutions.
Question 1.
x2 – x – 2 = 0
Answer:
The given equation is:
x² – x – 2 = 0
So,
The graph of the given equation in the coordinate plane is:

Now,
From the graph,
We can observe the solutions that satisfy the given equation i.e., x-intercepts from the graph are: -1 and 2
Now,
We have to check whether the given solutions satisfy or not algebraically
So,
So,
(-1)² – (-1) – 2 = 0
1 +1 – 2 = 0
2 – 2 = 0
0 = 0
(2)² – 2 + 2 = 0
4 – 2 + 2 = 0
2 – 2 = 0
0 = 0
Hence, from the above,
We can conclude that the solutions of the given equation are: -1 and 2

Question 2.
x2 + 7x = -10
Answer:
The given equation is:
x² + 7x = -10
x² + 7x + 10 = 0
So,
The graph of the given equation in the coordinate plane is:

Now,
From the graph,
We can observe the solutions that satisfy the given equation i.e., x-intercepts from the graph are: -2 and -5
Now,
We have to check whether the given solutions satisfy or not algebraically
So,
So,
(-2)² + 7(-2) + 10 = 0
4 – 14 + 10 = 0
14 – 14 = 0
0 = 0
(-5)² + 7(-5) + 10 = 0
25 – 35 + 10 = 0
35 – 35 = 0
0 = 0
Hence, from the above,
We can conclude that the solutions of the given equation are: -2 and -5

Question 3.
x2 + x = 12
Answer:
The given equation is:
x² + x = 12
x² + x – 12 = 0
So,
The graph of the given equation in the coordinate plane is:

Now,
From the graph,
We can observe the solutions that satisfy the given equation i.e., x-intercepts from the graph are: -4 and 3
Now,
We have to check whether the given solutions satisfy or not algebraically
So,
So,
(-4)² -4 – 12 = 0
16 – 4 – 12 = 0
12 – 12 = 0
0 = 0
(3)² + 3 – 12 = 0
9 – 12 + 3 = 0
12 – 12 = 0
0 = 0
Hence, from the above,
We can conclude that the solutions of the given equation are: -4 and 3

Solve the equation by graphing.
Question 4.
x2 + 36 = 12x
Answer:
The given equation is:
x²+ 36 = 12x
x² – 12x + 36 = 0
So,
The graph of the given equation in the coordinate plane is:

Now,
From the graph,
We can observe that there is only one x-intercept for the given equation
So,
That only x-intercept is the solution of the given equation
Hence, from the above,
We can conclude that the solution of the given equation is: 6

Question 5.
x2 + 4x = 0
Answer:
The given equation is:
x²+ 4x = 0
So,
The graph of the given equation in the coordinate plane is:

Now,
From the graph,
We can observe that there are two x-intercepts for the given equation
So,
That two x-intercepts are the solutions of the given equation
Hence, from the above,
We can conclude that the solutions of the given equation are: 0 and -4

Question 6.
x2 + 10x = -25
Answer:
The given equation is:
x²+ 10x = -25
x² + 10x + 25 = 0
So,
The graph of the given equation in the coordinate plane is:

Now,
From the graph,
We can observe that there is only one x-intercept for the given equation
So,
That one x-intercept is the solutions of the given equation
Hence, from the above,
We can conclude that the solution of the given equation is: -5

Question 7.
x2 = 3x – 3
Answer:
The given equation is:
x²= 3x – 3
x² – 3x + 3 = 0
So,
The graph of the given equation in the coordinate plane is:

Now,
From the graph,
We can observe that there are no x-intercepts for the given equation
So,
There are no real solutions for the given equation
Hence, from the above,
We can conclude that there are no real solutions for the given equation

Question 8.
x2 + 7x = -6
Answer:
The given equation is:
x²+ 7x = -6
x² + 7x + 6 = 0
So,
The graph of the given equation in the coordinate plane is:

Now,
From the graph,
We can observe that there are two x-intercepts for the given equation
So,
That two x-intercepts are the solutions of the given equation
Hence, from the above,
We can conclude that the solutions of the given equation are: -1 and -6

Question 9.
2x + 5 = -x2
Answer:
The given equation is:
2x + 5 = -x²
x² + 2x + 5 = 0
So,
The graph of the given equation in the coordinate plane is:

Now,
From the graph,
We can observe that there are no x-intercepts for the given equation
So,
There are no real solutions for the given equation
Hence, from the above,
We can conclude that there are no real solutions for the given equation

Question 10.
Graph f(x) = x2 + x – 6. Find the zeros of f.
Answer:
The given function is:
f(x) = x² + x – 6
To find the zeroes of f(x), make f(x) = 0
So,
x² + x – 6 = 0
x² + 3x – 2x – 6 = 0
x (x + 3) -2 (x + 3) = 0
(x – 2) (x + 3) = 0
x – 2 = 0 or x + 3 = 0
x = 2 or x = -3
Now,
The graph of f(x) in the coordinate plane is:

We know that,
If the equation is a 1-variable equation, then the intercepts of that variable are the solutions or the zeroes of the equation
Hence, from the above,
We can conclude that the zeroes of the given equation are: -3 and 2

Question 11.
Graph f(x) = -x2 + 2x + 2. Approximate the zeros of f to the nearest tenth.
Answer:

Question 12.
WHAT IF?
After how many seconds is the football 65 feet above the ground?
Answer:
From Example 6,
The function ‘h’ that represents the height (h) in feet of the football after t seconds is:
h = -16t² + 75t + 2
Now,
To determine when the football is above 65 feet from the ground,
Find the t-values for which h = 65
So,
-16t² + 75t + 2 = 65
-16t² + 75t + 2 – 65 = 0
-16t² + 75t – 63 = 0
Now,
The graph of the function h when the football is above 65 feet in the coordinate plane is:

So,
From the graph,
We can observe that the solutions i.e., t-intercepts for the given equation are two.
The first t-intercept is: 1
The second -intercept is in between 3 and 4
Hence, from the above,
We ca conclude that the football is 65 feet above the ground after 1 second and after about 3.8 seconds

Solving Quadratic Equations by Graphing 9.2 Exercises

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
What is a quadratic equation?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 1

Question 2.
WHICH ONE DOESN’T BELONG?
Which equation does not belong with the other three? Explain your reasoning.
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 4
Answer:
The given equations are:
a. x² + 5x = 20
b. x² + x – 4 = 0
c. x² – 6 = 4x
d. 7x + 12 = x²
So,
The graphs of the given equations in the coordinate plane are:

Now,
From the graph,
The solutions of the 4 equations are shown and we can observe that the three equations have the solutions whereas 1 equation does not have a real solution
Hence, from the graph,
We can conclude that x² – 7x + 12 = 0 does not belong with the other three

Question 3.
WRITING
How can you use a graph to find the number of solutions of a quadratic equation?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 3

Question 4.
WRITING
How are solutions, roots, x-intercepts, and zeros related?
Answer:
Zeroes, roots, and x-intercepts are all names for values that make a function equal to zero. That is, what values of x make the statement f(x) = 0 true. … That, geometrically, zeros, roots, or x-intercepts of a function f(x) are the values of x where the graph of the function f crosses the x-axis

Monitoring Progress and Modeling with Mathematics

In Exercises 5–8, use the graph to solve the equation.
Question 5.
-x2 + 2x + 3 = 0
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 5

Question 6.
x2 – 6x + 8 = 0
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 6
Answer:
The given equation is:
x² – 6x + 8 = 0
The graph of the given equation is:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 6
So,
From the graph,
We can observe that there are two x-intercepts i.e., two solutions for the  given equation
The two x-intercepts of the given equation are: 2 and 4
Hence, from the above,
We can conclude that the solutions of the given equation are: 2 and 4

Question 7.
x2 + 8x + 16 = 0
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 7
Answer:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 7

Question 8.
-x2 – 4x – 6 = 0
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 8
Answer:
The given equation is:
-x² – 4x – 6 = 0
x² + 4x + 6 = 0
The graph of the given equation is:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 8
So,
From the graph,
We can observe that there are no x-intercepts i.e., no real solutions for the  given equation
Hence, from the above,
We can conclude that there are no real solutions for the given equation

In Exercises 9–12, write the equation in standard form.
Question 9.
4x2 = 12
Answer:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 9

Question 10.
-x2 = 15
Answer:
The given equation is:
-x² = 15
So,
-x² – 15 = 0
– (x² + 15) = 0
x² + 15 = 0
Hence, from the above,
We can conclude that the given equation in standard form is:
-x² – 15 = 0 or x² + 15 = 0

Question 11.
2x – x2 = 1
Answer:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 11

Question 12.
5 + x = 3x2
Answer:
The given equation is:
5 + x = 3x²
So,
5 + x – 3x² = 0
– (3x² – x – 5) = 0
5 + x – 3x² = 0
Hence, from the above,
We can conclude that the given equation in the standard form is:
5 + x – 3x² = 0 or 5 + x – 3x² = 0

In Exercises 13–24, solve the equation by graphing.
Question 13.
x2 – 5x = 0
Answer:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 13

Question 14.
x2 – 4x + 4 = 0
Answer:
The given equation is:
x² – 4x + 4 = 0
So,
The graph of the given equation in the coordinate plane is:

Now,
From the graph,
We can observe that there is one x-intercept for the given equation
So,
That one x-intercept is the solution of the given equation
Hence, from the above,
We can conclude that the solution of the given equation is: 2

Question 15.
x2 – 2x + 5 = 0
Answer:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 15

Question 16.
x2 – 6x – 7 = 0
Answer:
The given equation is:
x² – 6x – 7 = 0
So,
The graph of the given equation in the coordinate plane is:

Now,
From the graph,
We can observe that there are two x-intercepts for the given equation
So,
That two x-intercepts are the solutions of the given equation
Hence, from the above,
We can conclude that the solutions of the given equation are: -1 and 7

Question 17.
x2 – 6x = 9
Answer:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 17

Question 18.
-x2 = 8x + 20
Answer:
The given equation is:
-x² =  8x + 20
x² + 8x + 20 = 0
So,
The graph of the given equation in the coordinate plane is:

Now,
From the graph,
We can observe that there are no x-intercepts for the given equation
So,
There are no real solutions for the given equation
Hence, from the above,
We can conclude that there are no real solutions for the given equation

Question 19.
x2 = -1 – 2x
Answer:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 19

Question 20.
x2 = -x – 3
Answer:
The given equation is:
x² = -x – 3
x² + x + 3 = 0
So,
The graph of the given equation in the coordinate plane is:

Now,
From the graph,
We can observe that there are no x-intercepts for the given equation
So,
There are no real solutions for the given equation
Hence, from the above,
We can conclude that there are no real solutions for the given equation

Question 21.
4x – 12 = -x2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 21

Question 22.
5x – 6 = x2
Answer:
The given equation is:
5x – 6 = x²
x² – 5x + 6 = 0
So,
The graph of the given equation in the coordinate plane is:

Now,
From the graph,
We can observe that there are two x-intercepts for the given equation
So,
That two x-intercepts are the solutions of the given equation
Hence, from the above,
We can conclude that the solutions of the given equation are: 2 and 3

Question 23.
x2 – 2 = -x
Answer:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 23

Question 24.
16 + x2 = -8x
Answer:
The given equation is:
16 + x² = -8x
x² + 8x + 16 = 0
So,
The graph of the given equation in the coordinate plane is:

Now,
From the graph,
We can observe that there is one x-intercept for the given equation
So,
That one x-intercept is the solution of the given equation
Hence, from the above,
We can conclude that the solution of the given equation is: -4

Question 25.
ERROR ANALYSIS
Describe and correct the error in solving x2 + 3x = 18 by graphing.
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 9
Answer:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 25

Question 26.
ERROR ANALYSIS
Describe and correct the error in solving x2 + 6x + 9 = 0 by graphing.
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 10
Answer:
The given equation in the standard form is:
x² + 6x + 9 = 0
So,
The graph of the given equation in the coordinate plane is:

Now,
From the graph,
We can observe that there is one x-intercept for the given equation
So,
That one x-intercept is the solution of the given equation
Hence, from the above,
We can conclude that the solution of the given equation is: x = -3

Question 27.
MODELING WITH MATHEMATICS
The height y (in yards) of a flop shot in golf can be modeled by y = -x2 + 5x, where x is the horizontal distance (in yards).
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 11
a. Interpret the x-intercepts of the graph of the equation.
b. How far away does the golf ball land?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 27

Question 28.
MODELING WITH MATHEMATICS
The height h (in feet) of an underhand volleyball serve can be modeled by h = -16t2 + 30t + 4, where t is the time (in seconds).
a. Do both t-intercepts of the graph of the function have meaning in this situation? Explain.
Answer:
It is given that
The height h (in feet) of an underhand volleyball serve can be modeled by
h = -16t² + 30t + 4
Where,
t is the time in seconds
Now,
The graph of the equation of h in the coordinate plane is:

Now,
From the graph,
We can observe that there are two t-intercepts.
So,
The two t-intercepts of the given equation are: 0 and 2
Now,
From the two intercepts of t,
We can observe that both the intercepts have the meaning
When t = 0,
The underhand volleyball can’t be served
When t = 2,
The underhand vollyball can be served
Hence, from the above,
We can conclude that the two intercepts of t have the meaning in this situation

b. No one receives the serve. After how many seconds does the volleyball hit the ground?
Answer:
From part (a),
We can observe that the t-intercepts are: 0 and 2
Now,
The time that the volleyball hit the ground = 2 – 0 = 2 seconds
Hence, from the above,
We can conclude that after 2 seconds, the volleyball hit the ground

In Exercises 29–36, solve the equation by using Method 2 from Example 3.
Question 29.
x2 = 10 – 3x
Answer:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 29

Question 30.
2x – 3 = x2
Answer:
The given equation is:
2x – 3 = x²
Let,
y = x²
y = 2x – 3
So,
The graph of the given equation in the coordinate plane is:

Now,
From the graph,
We can observe that the graphs of two equations do not intersect
Hence, from the above,
We can conclude that the given equation does not have any real solution

Question 31.
5x – 7 = x2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 31

Question 32.
x2 = 6x – 5
Answer:
The given equation is:
x² = 6x – 5
Let,
y = x²
y = 6x – 5
So,
The graph of the given equation in the coordinate plane is:

Now,
From the graph,
We can observe that the graphs of two equations intersect at (1, 1)
We know that,
As the given equation is in terms of x, the x-value in the intersection point of the two equations will become the solution of the given equation
Hence, from the above,
We can conclude that the solution of the given equation is: 1

Question 33.
x2 + 12x = -20
Answer:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 33

Question 34.
x2 + 8x = 9
Answer:
The given equation is:
x²  + 8x = 9
So,
x² = -8x + 9
Let,
y = x²
y = -8x + 9
So,
The graph of the given equation in the coordinate plane is:

Now,
From the graph,
We can observe that the graphs of two equations intersect at (1, 1)
We know that,
As the given equation is in terms of x, the x-value in the intersection point of the two equations will become the solution of the given equation
Hence, from the above,
We can conclude that the solution of the given equation is: 1

Question 35.
-x2 – 5 = -2x
Answer:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 35

Question 36.
-x2 – 4 = -4x
Answer:
The given equation is:
– x² – 4 = -4x
So,
-x² = -4x + 4
x² = 4x – 4
Let,
y = x²
y = 4x – 4
So,
The graph of the given equation in the coordinate plane is:

Now,
From the graph,
We can observe that the graphs of two equations intersect at (2, -4)
We know that,
As the given equation is in terms of x, the x-value in the intersection point of the two equations will become the solution of the given equation
Hence, from the above,
We can conclude that the solution of the given equation is: 2

In Exercises 37–42, find the zero(s) of f.
Question 37.
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 12
Answer:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 37

Question 38.
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 13
Answer:
The given function is:
f(x) = (x + 1) (x² + 6x + 8)
We know that,
To find the zeroes of a given function, we have to make the function equal to zero or when we use the graph of the given function, the x-intercepts are also considered as zeroes of the given function
So,
f(x) = 0
(x + 1) (x² + 6x + 8) = 0
x + 1 = 0 (or)  x² + 6x + 8 = 0
x + 1 = 0 (or)  x² + 4x + 2x + 8 = 0
x + 1 = 0 (or)  x (x + 4) + 2 (x + 4) = 0
x + 1 = 0 (or) (x + 2) (x + 4) = 0
x + 1 = 0 (or) x + 2 = 0 (or) x + 4 = 0
x = -1 (or) x = -2 (or) x = -4
Now,
From the given graph,
We can observe that the x-intercepts of the given graph are: -1, -2, and -4
Hence, from the above,
We can conclude that the zeroes of the given function are: -1, -2, and -4

Question 39.
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 14
Answer:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 39

Question 40.
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 15
Answer:
The given function is:
f(x) = (x – 5) (-x² + 3x – 3)
We know that,
To find the zeroes of a given function, we have to make the function equal to zero or when we use the graph of the given function, the x-intercepts are also considered as zeroes of the given function
So,
f(x) = 0
(x – 5) (-x² + 3x – 3) = 0
x – 5 = 0 (or)  -x² + 3x – 3 = 0
Now,
From
-x² + 3x – 3 = 0,
We can observe that there are no real solutions
So,
x – 5 = 0
x = 5
Now,
From the given graph,
We can observe that the x-intercept of the given graph is: -5
Hence, from the above,
We can conclude that the zero of the given function is: -5

Question 41.
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 16
Answer:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 41.1
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 41.2

Question 42
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 17
Answer:
The given function is:
f(x) = (x² + 1) (x² – x – 2)
We know that,
To find the zeroes of a given function, we have to make the function equal to zero or when we use the graph of the given function, the x-intercepts are also considered as zeroes of the given function
So,
f(x) = 0
(x² + 1) (x² – x – 2) = 0
x² + 1 = 0 (or)  x² – x – 2 = 0
x² + 1 = 0 (or)  x² + x – 2x – 2 = 0
x² + 1 = 0 (or)  x (x + 1) – 2 (x + 1) = 0
x² + 1 = 0 (or) (x + 1) (x – 2) = 0
x² + 1 = 0 (or) x + 1 = 0 (or) x – 2 = 0
x² = -1 (or) x = -1 (or) x = 2
No,
From
x² + 1 = 0,
We can observe that there are no real solutions
Now,
From the given graph,
We can observe that the x-intercepts of the given graph are: -1, and 2
Hence, from the above,
We can conclude that the zeroes of the given function are: -1, and 2

In Exercises 43–46, approximate the zeros of f to the nearest tenth.
Question 43.
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 18
Answer:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 43

Question 44.
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 19
Answer:

Question 45.
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 20
Answer:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 45

Question 46.
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 21
Answer:

In Exercises 47–52, graph the function. Approximate the zeros of the function to the nearest tenth, if necessary.
Question 47.
f(x) = x2 + 6x + 1
Answer:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 47.1
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 47.2

Question 48.
f(x) = x2 – 3x + 2
Answer:

Question 49.
y = -x2 + 4x – 2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 49.1
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 49.2

Question 50.
y = -x2 + 9x – 6
Answer:

Question 51.
f(x) = \(\frac{1}{2}\)x2 + 2x – 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 51.1
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 51.2

Question 52.
f(x) = -3x2 + 4x + 3
Answer:

Question 53.
MODELING WITH MATHEMATICS
At a Civil War reenactment, a cannonball is fired into the air with an initial vertical velocity of 128 feet per second. The release point is 6 feet above the ground. The function h = -16t2 + 128t + 6 represents the height h (in feet) of the cannonball after t seconds.
a. Find the height of the cannonball each second after it is fired.
b. Use the results of part (a) to estimate when the height of the cannonball is 150 feet.
c. Using a graph, after how many seconds is the cannonball 150 feet above the ground?
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 22
Answer:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 53.2

Question 54.
MODELING WITH MATHEMATICS
You throw a softball straight up into the air with an initial vertical velocity of 40 feet per second. The release point is 5 feet above the ground. The function h = -16t2 + 40t + 5 represents the height h (in feet) of the softball after t seconds.
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 23
a. Find the height of the softball each second after it is released.
Answer:
b. Use the results of part (a) to estimate when the height of the softball is 15 feet.
c. Using a graph, after how many seconds is the softball 15 feet above the ground?
Answer:

MATHEMATICAL CONNECTIONS In Exercises 55 and 56, use the given surface area S of the cylinder to find the radius r to the nearest tenth.
Question 55.
S = 225 ft2
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 24
Answer:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 55

Question 56.
S = 750 m2
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 25
Answer:
It is given that
The surface area of the cylinder (S) = 750 m²
We know that,
The surface area of a cylinder (S) = 2πr² + 2πrh
From the figure,
It is given that
The height of the cylinder (h) = 13 m
So,
750 = 2πr² + 2πrh
750 = 2πr² + 2πr (13)
750 = 2πr (r + 13)
375 = πr² + 13πr
πr² + 13πr – 375 = 0

From the above graph,
We can observe that the given equation has 2 roots and one is negative.
We know that,
The radius can’t be negative
Hence, from the above,
We can conclude that the radius of the given cylinder is: 6.21

Question 57.
WRITING
Explain how to approximate zeros of a function when the zeros are not integers.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 57

Question 58.
HOW DO YOU SEE IT?
Consider the graph shown.
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 26
a. How many solutions does the quadratic equation x2 = -3x + 4 have? Explain.
Answer:
The given equation is:
x² = -3x + 4
x² + 3x – 4 = 0
The graph of the given equation in the coordinate plane is:

From the graph,
We can observe that the x-intercepts of the given equation are: -4 and 1
Hence, from the above,
We can conclude that the solutions of the given equation are: -4 and 1

b. Without graphing, describe what you know about the graph of y = x2 + 3x – 4.
Answer:
The given equation is:
y = x² + 3x – 4
Now,
For the solutions of the given equation,
Put y = 0
So,
x² + 3x – 4 = 0
x² + 4x – x – 4 = 0
x (x + 4) – 1 (x + 4) = 0
(x – 1) (x + 4) = 0
x – 1 = 0 or x + 4 = 0
x = 1 or x = -4
Hence, from the above,
We can conclude that the solutions of the given equation are: -1 and 4

Question 59.
COMPARING METHODS
Example 3 shows two methods for solving a quadratic equation. Which method do you prefer? Explain your reasoning.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 59

Question 60.
THOUGHT PROVOKING
How many different parabolas have -2 and 2 as x-intercepts? Sketch examples of parabolas that have these two x-intercepts.
Answer:
There is only 1 parabola with -2 and 2 as x-intercepts
We know that,
If the equation has 2 solutions, then that equation will be a quadratic equation
So,
The equation with -2 and 2 as x-intercepts are:
(x+ 2) (x – 2) = 0
So,
The sketch of the parabola that has these 2 x-intercepts is:

Question 61.
MODELING WITH MATHEMATICS
To keep water off a road, the surface of the road is shaped like a parabola. A cross-section=[

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

of the road is shown in the diagram. The surface of the road can be modeled by y = -0.0017x2 + 0.041x, where x and y are measured in feet. Find the width of the road to the nearest tenth of a foot.
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 27.1
Answer:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 61

Question 62.
MAKING AN ARGUMENT
A stream of water from a fire hose can be modeled by y = -0.003x2 + 0.58x + 3,where x and y are measured in feet. A firefighter is standing 57 feet from a building and is holding the hose 3 feet above the ground. The bottom of a window of the building is 26 feet above the ground. Your friend claims the stream of water will pass through the window. Is your friend correct? Explain.
Answer:

REASONING In Exercises 63–65, determine whether the statement is always, sometimes, or never true. Justify your answer.
Question 63.
The graph of y = ax2 + c has two x-intercepts when a is negative.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 63

Question 64.
The graph of y = ax2 + c has no x-intercepts when a and c have the same sign.
Answer:
The given equation is:
y = ax² + c
Ex:
The graph of y = x² + 1 is:

Hence, from the graph,
We can conclude that the given statement is always false

Question 65.
The graph of y = ax2 + bx + c has more than two x-intercepts when a ≠ 0.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 65

Maintaining Mathematical Proficiency

Determine whether the table represents an exponential growth function, an exponential decay function, or neither. Explain.
Question 66.
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 27
Answer:
The given table is:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 27
So,
The representation of the given table in the coordinate plane is:

We know that,
If the graph is exponential, then the graph must gradually increase or gradually decrease
Hence, from the above,
We can conclude that the given table is exponential

Question 67.
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 28
Answer:
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations 9.2 a 67

Lesson 9.3 Solving Quadratic Equations Using Square Roots

Essential Question
How can you determine the number of solutions of a quadratic equation of the form ax2 + c = 0?
Answer:
Quadratic functions can have 0,1, or 2 solutions. This depends on the discriminant. Before getting to that, remember that the formula to find the solution of a second degree (quadratic) equation is given by:
x = (-b ±\(\sqrt{ b² – 4ac}\)/2a

EXPLORATION 1

The Number of Solutions of ax2 + c = 0
Work with a partner. Solve each equation by graphing. Explain how the number of solutions of ax2 + c = 0 relates to the graph of y = ax2 + c.
a. x2 – 4 = 0
Answer:
The given equation is:
x² – 4 = 0
Add with 4 on both sides
So,
x² – 4 + 4 = 0 + 4
x² = 4
√x² = √4
x = 2 or x = -2
The graph of the given equation in the coordinate plane is:

Hence, from the above,
We can conclude that there are 2 real solutions for the given equation i.e., -2 and 2

b. 2x2 + 5 = 0
Answer:
The given equation is:
2x² + 5 = 0
Subtract with 5 on both sides
So,
2x² + 5 – 5 = 0 – 5
2x² = -5
x² = –\(\frac{5}{2}\)
√x² = -√\(\frac{5}{2}\)
We know that,
“Square root” will not take any negative values
Now,
The graph of the given equation in the coordinate plane is:

Hence, from the above,
We can conclude that the given equation does not have any real solutions

c. x2 = 0
Answer:
The given equation is:
x² = 0
So,
√x² = √0
x = 0
Now,
The graph of the given equation in the coordinate plane is:

Hence, from the above,
We can conclude that the given equation has only one solution i.e., 0

d. x2 – 5 = 0
Answer:
The given equation is:
x² – 5 = 0
Add with 5 on both sides
So,
x² – 5 + 5 = 0 + 5
x² = 5
√x² = √5
x = √5 or x = -√5
So,
The graph of the given equation in the coordinate plane is:

Hence, from the above,
We can conclude that there are 2 real solutions i.e., √5 and -√5

EXPLORATION 2

Estimating Solutions
Work with a partner. Complete each table. Use the completed tables to estimate the solutions of x2 – 5 = 0. Explain your reasoning.
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 1
Answer:
The completed tables are:

The given equation is:
x² – 5 = 0
The solutions of the given equation will be obtained from the table when there is a sign change
Hence, from the above,
We can conclude that the solutions of the given equation are: 2.24 and -2.24

EXPLORATION 3

Using Technology to Estimate Solutions
Work with a partner. Two equations are equivalent when they have the same solutions.
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 2
a. Are the equations x2 – 5 = 0 and x2 = 5 equivalent? Explain your reasoning
Answer:
The given equation is:
x² – 5 = 0
Add with 5 on both sides
x² – 5 + 5= 0 + 5
x²= 5
Hence, from the above,
We can conclude that
x² – 5 and x² = 5 are equivalent

b. Use the square root key on a calculator to estimate the solutions of x2 – 5 = 0. Describe the accuracy of your estimates in Exploration 2.
Answer:
The given equation is:
x² – 5 = 0
Now,
The solutions of the given equation are:

Hence, from the above,
We can conclude that the solutions of the given equation are:
x = √5 or x = -√5

c. Write the exact solutions of x2 – 5 = 0.
Answer:
The given equation is:
x² – 5 = 0
From Exploration 2,
We can conclude that
The exact solutions of the given equation are: 2.24 and -2.24

Communicate Your Answer

Question 4.
How can you determine the number of solutions of a quadratic equation of the form ax2 + c = 0?
Answer:
Quadratic functions can have 0,1, or 2 solutions. This depends on the discriminant. Before getting to that, remember that the formula to find the solution of a second degree (quadratic) equation is given by:
x = (-b ±\(\sqrt{ b² – 4ac}\) / 2a

Question 5.
Write the exact solutions of each equation. Then use a calculator to estimate the solutions.
a. x2 – 2 = 0
The given equation is:
x² – 2 = 0
Now,

We know that,
The value of √2 is: 1.41
Hence, from the above,
We can conclude that
The exact solutions of the given equation are: 1.41 and -1.41

b. 3x2 – 18 = 0
Answer:
The given equation is:
3x² – 18 = 0
Now,

We know that,
The value of √6 is: 2.45
Hence, from the above,
We can conclude that the exact solutions of the given equation are: 2.45 and -2.45

c. x2 – 8 = 0
Answer:
The given equation is:
x² – 8 = 0
Now,
Add with 8 on both sides
x² – 8+ 8 = 0 + 8
x² = 8
√x² = √8
x = √8 or x = -√8
We know that,
The value of √8 is: 2.83
Hence, from the above,
We can conclude that the exact solutions of the given equation are: 2.83 and -2.83

Monitoring Progress

Solve the equation using square roots.
Question 1.
-3x2 = -75
Answer:
The given equation is:
-3x² = -75
3x² = 75
x² = \(\frac{75}{3}\)
x² = 25
√x² = √25
x = 5 or x = -5
Hence, from the above,
We can conclude that
The solutions of the given equation are: 5 and -5

Question 2.
x2 + 12 = 10
Answer:
The given equation is:
x² + 12 = 10
x² = 10 – 12
x² = -2
√x² = √-2
We know that,
The Square root does not take any negative values
Hence, from the above,
We can conclude that the given equation does not have any real solutions

Question 3.
4x2 – 15 = -15
Answer:
The given equation is:
4x² – 15 = -15
4x² = -15 + 15
4x² = 0
x² = 0
√x² = √0
x = 0
Hence, from the above,
We can conclude that the solution of the given equation is: 0

Question 4.
(x + 7)2 = 0
Answer:
The given equation is:
(x + 7)² = 0
√(x + 7)² = √0
x + 7 = 0
x = 0 – 7
x = -7
Hence, from the above,
We can conclude that the solution of the given equation is: -7

Question 5.
4(x – 3)2 = 9
Answer:
The given equation is:
4 (x – 3)² = 9
(x – 3)² = \(\frac{9}{4}\)
√(x – 3)² = √\(\frac{9}{4}\)
x – 3 = \(\frac{3}{2}\) or x – 3 = –\(\frac{3}{2}\)
x = \(\frac{3}{2}\) + 3 or x = –\(\frac{3}{2}\) + 3
x = \(\frac{9}{2}\) or x = \(\frac{3}{2}\)
Hence, from the above,
We can conclude that the solutions of the given equation are: \(\frac{9}{2}\) and \(\frac{3}{2}\)

Question 6.
(2x + 1)2 = 36
Answer:
The given equation is:
(2x + 1)² = 36
√(2x + 1)² = √36
2x + 1 = 6 or 2x + 1 = -6
2x = 6 – 1 or 2x = -6 – 1
2x = 5 or 2x = -7
x = \(\frac{5}{2}\) or x = –\(\frac{7}{2}\)
Hence, from the above,
We can conclude that the solutions of the given equation are: \(\frac{5}{2}\) and –\(\frac{7}{2}\)

Solve the equation using square roots. Round your solutions to the nearest hundredth.
Question 7.
x2 + 8 = 19
Answer:
The given equation is:
x² + 8 = 19
Subtract with 8 on both sides
x² + 8 – 8 = 19 – 8
x² = 11
√x² = √11
x = √11 or x = -√11
We know that,
The value of √11 is: 3.31
Hence, from the above,
We can conclude that the solutions of the given equation are: 3.31 and -3.31

Question 8.
5x2 – 2 = 0
Answer:
The given equation is:
5x² – 2 = 0
Add with 2 on both sides
5x² – 2 + 2 = 0 + 2
5x² = 2
x² = \(\frac{2}{5}\)
√x² = √\(\frac{2}{5}\)
x = √\(\frac{2}{5}\) or x = -√\(\frac{2}{5}\)
We know that,
The value of \(\frac{2}{5}\) is: 0.40
Hence, from the above,
We can conclude that the solutions of the given equation are: 0.40 and -0.40

Question 9.
3x2 – 30 = 4
Answer:
The given equation is:
3x² – 30 = 4
3x² = 4 + 30
3x²= 34
x² = \(\frac{34}{3}\)
√x² = √\(\frac{34}{3}\)
x = √\(\frac{34}{3}\) or x = -√\(\frac{34}{3}\)
We know that,
The value of \(\frac{34}{3}\) is: 11.33
Hence, from the above,
We can conclude that the solutions of the given equation are: 11.33 ad -11.33

Question 10.
WHAT IF?
In Example 4, the volume of the tank is 315 cubic feet. Find the length and width of the tank.
Answer:
From Example 4,
It is given that
A touch tank has a height of 3 feet and its length is 3 times its width
We know that,
The volume of the tank (V) = l × w × h
315 = 3w (w) (3)
315 = 9w²
w² = \(\frac{315}{9}\)
w² = 35
√w² = √35
w = √35 or w = -√35
We know that,
The volume won’t be negative
The value of √35 is: 5.91
So,
The width of the tank is: 5.91
The length of the tank is: 3 (5.91) = 17.74
Hence, from the above,
We can conclude that
The length of the tank is: 17.74
The width of the tank is 5.91

Question 11.
The surface area S of a sphere with radius r is given by the formula S = 4πr2. Solve the formula for r. Then find the radius of a globe with a surface area of 804 square inches.
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 3
Answer:
It is given that
The surface area S of a sphere is given by the formula
S = 4πr²
So,
r² = \(\frac{S}{4π}\)
√r² = √\(\frac{S}{4π}\)
r = √\(\frac{S}{4π}\) or r = -√\(\frac{S}{4π}\)
We know that,
The surface area S of the sphere won’t take negative values
So,
r = √\(\frac{S}{4π}\)
So,
r = √\(\frac{804}{4π}\)
r = 64.01 units
Hence, from the above,
We can conclude that the radius of a globe with a surface area of 804 sq. units is: 64.01 units

Solving Quadratic Equations Using Square Roots 9.3 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
The equation x2 = d has ____ real solutions when d > 0.
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 a 1

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 4
Answer:
The given equations are:
a. x² = 144
b. x² – 144 =0
c. x² + 146 = 2
d. x² + 2 = 146
Now,
From the above 4 equations,
We can observe that
a, b, and d are the same because in these 3 equations,
x² = 144
But, in d,
x² = -144
Hence, from the above,
We can conclude that the equations a, b, and d are the same whereas the equation c is different

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, determine the number of real solutions of the equation. Then solve the equation using square roots.
Question 3.
x2 = 25
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 a 3

Question 4.
x2 = -36
Answer:
The given equation is:
x² = -36
√x² = √-36
x = √-36
We know that,
The square root won’t take the negative values
Hence, from the above,
We can conclude that there are no real solutions for the given equation

Question 5.
x2 = -21
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 a 5

Question 6.
x2 = 400
Answer:
The given equation is:
x² = 400
√x² = √400
x = 20 or x = -20
Hence, from the above,
We can conclude that there are 2 real solutions for the given equation i.e., 20 and -20

Question 7.
x2 = 0
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 a 7

Question 8.
x2 = 169
Answer:
The given equation is:
x² = 169
√x² = √169
x = 13 or x = -13
Hence, from the above,
We can conclude that there are 2 real solutions for the given equation i.e., 13 and -13

In Exercises 9–18, solve the equation using square roots.
Question 9.
x2 – 16 = 0
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 a 9

Question 10.
x2 + 6 = 0
Answer:
The given equation is:
x² + 6 = 0
Subtract with  on both sides
x² + 6 – 6 = 0 – 6
x² = -6
√x² = √-6
x = √-6
We know that,
The square root won’t take any negative values
Hence, from the above,
We can conclude that there are no real solutions for the given equation

Question 11.
3x2 + 12 = 0
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 a 11

Question 12.
x2 – 55 = 26
Answer:
The given equation is:
x² – 55 = 26
Add with 55 on both sides
x² – 55 + 55 = 26 + 55
x² = 81
√x² = √81
x = 9 or x = -9
Hence, from the above,
We can conclude that
The solutions of the given equation are: 9 and -9

Question 13.
2x2 – 98 = 0
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 a 13

Question 14.
-x2 + 9 = 9
Answer:
The given equation is:
-x² + 9 = 9
Subtract with 9 on both sides
-x² + 9 – 9 = 9 – 9
-x² = 0
x² = 0
√x² = √0
x = 0
Hence, from the above,
We can conclude that
The solution of the given equation is:  0

Question 15.
-3x2 – 5 = -5
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 a 15

Question 16.
4x2 – 371 = 29
Answer:
The given equation is:
4x² – 371 = 29
Add with 371 on both sides
4x² – 371 + 371 = 29 + 371
4x² = 400
x² = \(\frac{400}{4}\)
x² = 100
√x² = √100
x = 10 or x = -10
Hence, from the above,
We can conclude that
The solutions of the given equation are: 10 and -10

Question 17.
4x2 + 10 = 11
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 a 17

Question 18.
9x2 – 35 = 14
Answer:
The given equation is:
9x² – 35 = 14
Add with 35 on both sides
9x² – 35 + 5 = 14 + 35
9x² = 49
x² = \(\frac{49}{9}\)
√x² = √\(\frac{49}{9}\)
x = \(\frac{7}{3}\) or x = –\(\frac{7}{3}\)
Hence, from the above,
We can conclude that
The solutions of the given equation are: \(\frac{7}{3}\) and –\(\frac{7}{3}\)

In Exercises 19–24, solve the equation using square roots.
Question 19.
(x + 3)2 = 0
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 a 19

Question 20.
(x – 1)2 = 4
Answer:
The given equation is:
(x – 1)² = 4
√(x – 1)² = √4
x – 1 = 2 or x – 1 = -2
x = 2 + 1 or x = -2 + 1
x = 3 or x = -1
Hence, from the above,
We can conclude that
The solutions of the given equation are: 3 and -1

Question 21.
(2x – 1)2 = 81
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 a 21

Question 22.
(4x + 5)2 = 9
Answer:
The given equation is:
(4x + 5)² = 9
√(4x + 5)² = √9
4x + 5 = 3 or 4x + 5 = -3
4x = 3 – 5 or 4x = -3 – 5
4x = -2 or 4x = -8
x = \(\frac{-2}{4}\) or x = \(\frac{-8}{4}\)
x = –\(\frac{1}{2}\) or x = -2
Hence, from the above,
We can conclude that
The solutions of the given equation are: –\(\frac{1}{2}\) and -2

Question 23.
9(x + 1)2 = 16
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 a 23

Question 24.
4(x – 2)2 = 25
Answer:
The given equation is:
4 (x – 2)² = 25
(x – 2)² = \(\frac{25}{4}\)
√(x – 2)² = √\(\frac{25}{4}\)
x – 2 = \(\frac{5}{2}\) or x – 2 = –\(\frac{5}{2}\)
x= \(\frac{5}{2}\) + 2 or x = –\(\frac{5}{2}\) + 2
x = \(\frac{9}{2}\) or x = –\(\frac{1}{2}\)
Hence, from the above,
We can conclude that
The solutions of the given equation are: \(\frac{9}{2}\) and –\(\frac{1}{2}\)

In Exercises 25–30, solve the equation using square roots. Round your solutions to the nearest hundredth.
Question 25.
x2 + 6 = 13
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 a 25

Question 26.
x2 + 11 = 24
Answer:
The given equation is:
x² + 11 = 24
Subtract with 11 on both sides
x² + 11 – 11 = 24 – 11
x² = 13
√x² = √13
x = √13 or x = -√13
We know that,
√13 = 3.60
Hence, from the above,
We can conclude that the solutions of the given equation are: 3.60 and -3.60

Question 27.
2x2 – 9 = 11
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 a 27

Question 28.
5x2 + 2 = 6
Answer:
The given equation is:
5x² + 2 = 6
Subtract with 2 on both sides
5x² + 2 – 2 = 6 – 2
5x² = 4
x² = \(\frac{4}{5}\)
√x² = √\(\frac{4}{5}\)
x = \(\frac{2}{√5}\) or x = –\(\frac{2}{√5}\)
We know that,
√5 = 2.23
So,
\(\frac{2}{√5}\) = 0.89
Hence, from the above,
We can conclude that the solutions of the given equation are: 0.89 and -0.89

Question 29.
-21 = 15 – 2x2
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 a 29

Question 30.
2 = 4x2 – 5
Answer:
The given equation is:
2 = 4x² – 5
Add with 5 on both sides
2 + 5 = 4x² – 5 + 5
7 = 4x²
x² = \(\frac{7}{4}\)
√x² = √\(\frac{7}{4}\)
x = √\(\frac{7}{4}\) or x = -√\(\frac{7}{4}\)
We know that,
√\(\frac{7}{4}\) = 1.32
Hence, from the above,
We can conclude that the solutions of the given equation are: 1.32 and -1.32

Question 31.
ERROR ANALYSIS
Describe and correct the error in solving the equation 2x2 – 33 = 39 using square roots.
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 5
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 a 31

Question 32.
MODELING WITH MATHEMATICS
An in-ground pond has the shape of a rectangular prism. The pond has a depth of 24 inches and a volume of 72,000 cubic inches. The length of the pond is two times its width. Find the length and width of the pond.
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 6
Answer:
It is given that a pond is in the shape of a rectangular prism
We know that,
The volume of a rectangular prism (V) = l × w × h
It is also given that the length of the pond is two times its width
So,
V = (2w) × w × h
72,000 = 2w² × 24
2w² = \(\frac{72,000}{24}\)
2w² = 3,000
w² = \(\frac{3,000}{2}\)
w² = 1,500
√w² = √1,500
w = √1,500 or w = -√1,500
We know that,
The width will not take any negative values
So,
w = √1,500
We know that,
√1,500 = 38.72 inches
So,
The width of the rectangular prism = 38.72 inches
The length of the rectangular prism = 2 (38.72) = 77.45 inches
Hence, from the above,
We can conclude that
The length of the rectangular prism is: 77.45 inches
The width of the rectangular prism is: 38.72 inches

Question 33.
MODELING WITH MATHEMATICS
A person sitting in the top row of the bleachers at a sporting event drops a pair of sunglasses from a height of 24 feet. The function h = -16x2 + 24 represents the height h (in feet) of the sunglasses after x seconds. How long does it take the sunglasses to hit the ground?
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 a 33

Question 34.
MAKING AN ARGUMENT
Your friend says that the solution of the equation x2 + 4 = 0 is x = 0. Your cousin says that the equation has no real solutions. Who is correct? Explain your reasoning.
Answer:
Your cousin is correct

Explanation:
The given equation is:
x² + 4 = 0
Subtract with 4 on both sides
x² + 4 – 4 = 0- 4
x² = -4
We know that,
The square of a real number can’t take any negative numbers
So,
There are no real solutions for the given equation
Hence, from the above,
We can conclude that your cousin is correct

Question 35.
MODELING WITH MATHEMATICS
The design of a square rug for your living room is shown. You want the area of the inner square to be 25% of the total area of the rug. Find the side length x of the inner square.
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 7
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 a 35

Question 36.
MATHEMATICAL CONNECTIONS
The area A of a circle with radius r is given by the formula A = πr2.
a. Solve the formula for r.
Answer:
It is given that
The area A of a circle with radius r is given as:
A = πr²
r² = \(\frac{A}{π}\)
√r² = √\(\frac{A}{π}\)
r = √\(\frac{A}{π}\) or r = -√\(\frac{A}{π}\)
We know that,
The radius of the circle won’t take any negative values
Hence,
r = √\(\frac{A}{π}\)

b. Use the formula from part (a) to find the radius of each circle.
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 8
Answer:
From part (a),
r = √\(\frac{A}{π}\)
For A = 113ft²,
r = √\(\frac{113}{π}\)
r = 5.99ft
For A = 810 in.²,
r = √\(\frac{1810}{π}\)
r = 24 in.
For A = 531m²,
r = √\(\frac{531}{π}\)
r = 13m

c. Explain why it is beneficial to solve the formula for r before finding the radius.
Answer:
It is beneficial to solve the formula for r before finding the radius because the formula for r is complicated due to square root and it will become easy if we simplified the formula for r before finding the value of r

Question 37.
WRITING
How can you approximate the roots of a quadratic equation when the roots are not integers?
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 a 37

Question 38.
WRITING
Given the equation ax2 + c = 0, describe the values of a and c so the equation has the following number of solutions.
a. two real solutions
b. one real solution
c. no real solutions
Answer:
The given equation is:
ax² + c = 0
Subtract with c on both sides
ax²  + c – c = 0 – c
ax² = -c
Now,
a. Two real solutions
For the equations to have 2 real solutions,
a —-> Positive
c —–> Negative
b. One real solution
For the equation to have 1 real solution,
a —–> Positive
c ——> Zero
c. No real solutions
For the equation to have no real solutions,
a —–> Positive
c —–> Positive

Question 39.
REASONING
Without graphing, where do the graphs of y = x2 and y = 9 intersect? Explain.
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 a 39

Question 40.
HOW DO YOU SEE IT?
The graph represents the function f(x) = (x – 1)2. How many solutions does the equation (x – 1)2 = 0 have? Explain.
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 9
Answer:
The given equation is:
f (x) = (x – 1)²
Now,
It is given that
f (x) = 0
So,
(x – 1)² = 0
√(x – 1)² = √0
x – 1 = 0
x = 0 + 1
x = 1
Hence, from the above,
We can conclude that there is only 1 solution for the given equation i.e., 1

Question 41.
REASONING
Solve x2 = 1.44 without using a calculator. Explain your reasoning.
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 a 41

Question 42.
THOUGHT-PROVOKING
The quadratic equation ax2 + bx + c = 0 can be rewritten in the following form. \(\left(x+\frac{b}{2 a}\right)^{2}=\frac{b^{2}-4 a c}{4 a^{2}}\) Use this form to write the solutions of the equation.
Answer:

Question 43.
REASONING
An equation of the graph shown is y = \(\frac{1}{2}\)(x – 2)2 + 1. Two points on the parabola have y-coordinates of 9. Find the x-coordinates of these points.
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 10
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 a 43

Question 44.
CRITICAL THINKING
Solve each equation without graphing.
a. x2 – 12x + 36 = 64
The given equation is:
x² – 12x + 36 = 64

Hence, from the above,
We can conclude that
The solutions of the given equation are: -2 and 14

b. x2 + 14x + 49 = 16
Answer:
The given equation is:
x² + 14x + 49 = 16

Hence, from the above,
We can conclude that
The solutions for the given equation are: -3 and -11

Maintaining Mathematical Proficiency

Factor the polynomial.
Question 45.
x2 + 8x + 16
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 a 45

Question 46.
x2 – 4x + 4
Answer:
The given expression is:
x² – 4x + 4
So,
x² – 4x + 4
= x² – 2x – 2x + 4
= x (x – 2) – 2 (x – 2)
= (x – 2) ( x – 2)
= (x – 2)²
Hence, from the above,
We can conclude that
The factor of the given polynomial is: (x – 2)²

Question 47.
x2 – 14x + 49
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 a 47

Question 48.
x2 + 18x + 81
Answer:
The given expression is:
x² + 18x + 81
So,
x² + 18x + 81
= x² + 9x + 9x + 81
= x (x + 9) + 9 (x + 9)
= (x + 9) (x + 9)
= (x + 9)²
Hence, from the above,
We can conclude that
The factor of the given polynomial is: (x + 9)²

Question 49.
x2 + 12x + 36
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 a 49

Question 50.
x2 – 22x + 121
Answer:
The given expression is:
x² – 22x + 121
So,
x² – 22x + 121
= x² – 11x – 11x + 121
= x (x – 11) – 11 (x – 11)
= (x – 11) (x – 11)
= (x – 11)²
Hence, from the above,
We can conclude that
The factor of the given polynomial is: (x – 11)²

Solving Quadratic Equations Study Skills: Keeping a Positive Attitude

9.1–9.3 What Did You Learn?

Core Vocabulary
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 11

Core Concepts
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 12

Mathematical Practices

Question 1.
For each part of Exercise 100 on page 488 that is sometimes true, list all examples and counterexamples from the table that represent the sum or product being described.
Answer:

Question 2.
Which Examples can you use to help you solve Exercise 54 on page 496?
Answer:
You can use Example 6 on page 493 to solve Exercise 54 on page 496

Question 3.
Describe how solving a simpler equation can help you solve the equation in Exercise 41 on page 502.
Answer:
The equation in Exercise 41 on page 502 is in the form of:
x² – a = 0
We know that,
If d> 0 i.e., the square is a positive number, then the only square root exists
Hence, in the above way,
We can solve Exercise 41 on page 502

Study Skills: Keeping a Positive Attitude

Do you ever feel frustrated or overwhelmed by math? You’re not alone. Just take a deep breath and assess the situation. Try to find a productive study environment, review your notes and the examples in the textbook, and ask your teacher or friends for help.
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations 9.3 13

Solving Quadratic Equations 9.1 – 9.3 Quiz

Simplify the expression.
Question 1.
\(\sqrt{112 x^{3}}\)
Answer:
The given expression is:
\(\sqrt{112 x^{3}}\)
We know that,
\(\sqrt{{a}\cdot b}\) = \(\sqrt{a}\) ⋅ \(\sqrt{b}\)
So,
\(\sqrt{112 x^{3}}\)
= \(\sqrt{{112 x^{2}}\cdot x}\)
= \(\sqrt{16 × 7 x^{2}}\) ⋅ \(\sqrt{x}\)
= 4x \(\sqrt{7x}\)
Hence, from the above,
We can conclude that
\(\sqrt{112 x^{3}}\) = 4x \(\sqrt{7x}\)

Question 2.
\(\sqrt{\frac{18}{81}}\)
Answer:
The given expression is:
\(\sqrt{\frac{18}{81}}\)
We know that,
\(\frac{\sqrt{a}}{\sqrt{b}}\) = \(\sqrt{\frac{a}{b}}\)
So,
\(\sqrt{\frac{18}{81}}\)
= \(\sqrt{\frac{18}{81}}\)
= \(\frac{\sqrt{18}}{\sqrt{81}}\)
= \(\frac{3\sqrt{2}}{9}\)
= \(\frac{\sqrt{2}}{3}\)
Hence, from the above,
We can conclude that
\(\sqrt{\frac{18}{81}}\) = \(\frac{\sqrt{2}}{3}\)

Question 3.
\(\sqrt[3]{-625}\)
Answer:
The given expression is:
\(\sqrt[3]{-625}\)
We know that,
\(\sqrt[3]{a}{b}\) = \(\sqrt[3]{a}\) ⋅ \(\sqrt[3]{b}\)
So,
\(\sqrt[3]{-625}\)
= \(\sqrt[3]{-625}{1}\)
= \(\sqrt[3]{-625}\) ⋅ \(\sqrt[3]{1}\)
= \(\sqrt[3]{-125 × 5}\)
= \(\sqrt[3]{-125}\) ⋅ \(\sqrt[3]{5}\)
= -5\(\sqrt[3]{5}\)
Hence, from the above,
We can conclude that
\(\sqrt[3]{-625}\) = -5\(\sqrt[3]{5}\)

Question 4.
\(\frac{12}{\sqrt{32}}\)
Answer:
The given expression is:
\(\frac{12}{\sqrt{32}}\)
We have to rationalize the denominator to simplify the given expression
So,
To rationalize the denominator, multiply and divide the given fraction with \(\sqrt{32}\)
So,
\(\frac{12}{\sqrt{32}}\)
= \(\frac{12}{\sqrt{32}}\) × (\(\sqrt{32}\) / \(\sqrt{32}\))
= \(12\sqrt{32}\) / (\(\sqrt{32}\) × \(\sqrt{32}\))
= \(12\sqrt{32}\) / 32
= \(3\sqrt{32}\) / 8
Hence, from the above,
We can conclude that
\(\frac{12}{\sqrt{32}}\) = \(3\sqrt{32}\) / 8

Question 5.
\(\frac{4}{\sqrt{11}}\)
Answer:
The given expression is:
\(\frac{4}{\sqrt{11}}\)
We have to rationalize the denominator to simplify the given expression
So,
To rationalize the denominator, multiply and divide the given fraction with \(\sqrt{11}\)
So,
\(\frac{4}{\sqrt{11}}\)
= \(\frac{4}{\sqrt{11}}\) × (\(\sqrt{11}\) / \(\sqrt{11}\))
= \(4\sqrt{11}\) / (\(\sqrt{11}\) × \(\sqrt{11}\))
= \(4\sqrt{11}\) / 11
Hence, from the above,
We can conclude that
\(\frac{4}{\sqrt{11}}\) = \(4\sqrt{11}\) / 11

Question 6.
\(\sqrt{\frac{144}{13}}\)
Answer:
The given expression is:
\(\sqrt{\frac{144}{13}}\)
We know that,
\(\sqrt{\frac{a}{b}}\) = \(\frac{\sqrt{a}}{\sqrt{b}}\)
So,
\(\sqrt{\frac{144}{13}}\)
= \(\frac{\sqrt{144}}{\sqrt{13}}\)
We have to rationalize the denominator to simplify the given expression
So,
To rationalize the denominator, multiply and divide the given fraction with \(\sqrt{13}\)
So,
\(\frac{\sqrt{144}}{\sqrt{13}}\)
= \(\frac{\sqrt{144}}{\sqrt{13}}\) × (\(\sqrt{13}\) / \(\sqrt{13}\))
= \(\sqrt{144 × 13}\) / (\(\sqrt{13}\) × \(\sqrt{13}\))
= 12\(\sqrt{13}\) / 13
Hence, from the above,
We can conclude that
\(\sqrt{\frac{14}{13}}\) = 12\(\sqrt{13}\) / 13

Question 7.
\(\sqrt[3]{\frac{54 x^{4}}{343 y^{6}}}\)
Answer:
The given expression is:
\(\sqrt[3]{\frac{54 x^{4}}{343 y ^{6}}}\)
We know that,
\(\sqrt[3]{\frac{a}{b}}\) = \(\frac{\sqrt[3]{a}}{\sqrt[3]{b}}\)
So,
\(\sqrt[3]{\frac{54 x^{4}}{343 y^{6}}}\)
= \(\frac{\sqrt[3]{54 x^{4}}}{\sqrt[3]{343 y^{6}}}\)
= \(\frac{\sqrt[3]{27 x^{3} × 2x}}{\sqrt[3]{343 y^{6}}}\)
= \(\frac{3x\sqrt[3]{ 2x}}{7 y^{2}}\)
Hence, from the above,
We can conclude that
\(\sqrt[3]{\frac{54 x^{4}}{343 y^{6}}}\) = \(\frac{3x\sqrt[3]{ 2x}}{7 y^{2}}\)

Question 8.
\(\sqrt{\frac{4 x^{2}}{28 y^{4} z^{5}}}\)
Answer:
The given expression is:
\(\sqrt{\frac{4 x^{2}}{28 y^{4} z^{5}}}\)
We know that,
\(\sqrt{\frac{a}{b}}\) = \(\frac{\sqrt{a}}{\sqrt{b}}\)
So,
\(\sqrt{\frac{4 x^{2}}{28 y^{4} z^{5}}}\)
= \(\frac{\sqrt{4 x^{2}}}{\sqrt{28 y^{4} z^{5}}}\)
= \(\frac{x}{y^{2} z^{2}\sqrt{7z}}\)
Hence, from the above,
We can conclude that
\(\sqrt{\frac{4 x^{2}}{28 y^{4} z^{5}}}\) = \(\frac{x}{y^{2} z^{2}\sqrt{7z}}\)

Question 9.
\(\frac{6}{5+\sqrt{3}}\)
Answer:
The given expression is:
\(\frac{6}{5 + \sqrt{3}}\)
We have to rationalize the denominator to simplify the given expression
So,
To rationalize the denominator, multiply and divide the given fraction with \(5 – \sqrt{3}\)
So,
\(\frac{6}{5 + \sqrt{3}}\)
= \(\frac{6}{5 + \sqrt{3}}\) × (\(5 – \sqrt{3}\) / \(5 – \sqrt{3}\))
= 6\((5 – \sqrt{3})\) / (\(5 + \sqrt{3}\) × \(5 – \sqrt{3}\))
= 6\((5 – \sqrt{3})\) / 22
= 6\((5 – \sqrt{3})\) / 22
Hence, from the above,
We can conclude that
\(\frac{6}{5 + \sqrt{3}}\) = 6 \((5 – \sqrt{3})\) / 22

Question 10.
2\(\sqrt{5}\) + 7\(\sqrt{10}\) – 3\(\sqrt{20}\)
Answer:
The given expression is:
2\(\sqrt{5}\) + 7\(\sqrt{10}\) – 3\(\sqrt{20}\)
So,
2\(\sqrt{5}\) + 7\(\sqrt{10}\) – 3\(\sqrt{20}\)
= 2\(\sqrt{5}\) + 7\(\sqrt{10}\) – 3\(\sqrt{5 × 4}\)
= 2\(\sqrt{5}\) + 7\(\sqrt{10}\) – 3 (2)\(\sqrt{5}\)
= 2\(\sqrt{5}\) + 7\(\sqrt{10}\) – 6\(\sqrt{5}\)
= \(\sqrt{5}\) ( 2 – 6) + 7\(\sqrt{10}\)
= \(\sqrt{5}\) (-4) + 7\(\sqrt{10}\)
= -4\(\sqrt{5}\) + 7\(\sqrt{10}\)
Hence, from the above,
We can conclude that
2\(\sqrt{5}\) + 7\(\sqrt{10}\) – 3\(\sqrt{20}\) = -4\(\sqrt{5}\) + 7\(\sqrt{10}\)

Question 11.
\(\frac{10}{\sqrt{8}-\sqrt{10}}\)
Answer:
The given expression is:
\(\frac{10}{\sqrt{8}-\sqrt{10}}\)
We have to rationalize the denominator to simplify the given expression
So,
To rationalize the denominator, multiply and divide the given fraction with \(\sqrt{8} +\sqrt{10} \)
So,
\(\frac{10}{\sqrt{8}-\sqrt{10}}\)
= \(\frac{10}{\sqrt{8}-\sqrt{10}}\) × (\(\sqrt{8} +\sqrt{10} \) / \(\sqrt{8} +\sqrt{10} \))
= 10(\(\sqrt{8} +\sqrt{10} \)) / (\(\sqrt{8} -\sqrt{10} \) × \(\sqrt{8} +\sqrt{10} \))
= 10(\(\sqrt{8} +\sqrt{10}\)) / -2
= – 10(\(\sqrt{8} +\sqrt{10} \))/ 2
Hence, from the above,
We can conclude that
\(\frac{10}{\sqrt{8}-\sqrt{10}}\) = – 10(\(\sqrt{8} +\sqrt{10} \))/ 2

Question 12.
\(\sqrt{6}\)(7\(\sqrt{12}\) – 4\(\sqrt{3}\))
Answer:
The given expression is:
\(\sqrt{6}\)(7\(\sqrt{12}\) – 4\(\sqrt{3}\))
So,
\(\sqrt{6}\)(7\(\sqrt{12}\) – 4\(\sqrt{3}\))
= \(\sqrt{6}\)(-4\(\sqrt{3}\) + 7\(\sqrt{4 × 3}\))
= \(\sqrt{6}\)(-4\(\sqrt{3}\) + 7 (2)\(\sqrt{3}\))
= \(\sqrt{6}\)(-4\(\sqrt{3}\) + 14\(\sqrt{3}\))
= \(\sqrt{6}\)(\(\sqrt{3}\) (14 – 4))
= \(\sqrt{6}\)(\(\sqrt{3}\) (12))
= 12(\(\sqrt{3}\))(\(\sqrt{6}\) )
= 12\(\sqrt{18}\)
= 12\(\sqrt{9 × 2}\)
= 36\(\sqrt{2}\)
Hence, from the above,
We can conclude that
\(\sqrt{6}\)(7\(\sqrt{12}\) – 4\(\sqrt{3}\)) = 36\(\sqrt{2}\)

Use the graph to solve the equation.
Question 13.
x2 – 2x – 3 = 0
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations q 1
Answer:
We know that,
The solutions of the equations are the x-intercepts of those equations when we draw the graph for them
Hence, from the above,
We can conclude that
The solutions of the given equation are: -1 and 3

Question 14.
x2 – 2x + 3 = 0
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations q 2
Answer:
We know that,
The solutions of the equations are the x-intercepts of those equations when we draw the graph for them
Hence, from the above,
We can conclude that
There are no real solutions for the given equation

Question 15.
x2 + 10x + 25 = 0
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations q 3
Answer:
We know that,
The solutions of the equations are the x-intercepts of those equations when we draw the graph for them
Hence, from the above,
We can conclude that
The solution of the given equation is: -5

Solve the equation by graphing.
Question 16.
x2 + 9x + 14 = 0
Answer:
The given equation is:
x² + 9x + 14 = 0
The representation of the given equation in the coordinate plane is:

We know that,
The solutions of the equations are the x-intercepts of those equations when we draw the graph for them
Hence, from the above,
We can conclude that
The solutions of the given equation are: -2 and -7

Question 17.
x2 – 7x = 8
Answer:
The given equation is:
x² – 7x = 8
So,
x² – 7x – 8 = 0
The representation of the given equation in the coordinate plane is:

We know that,
The solutions of the equations are the x-intercepts of those equations when we draw the graph for them
Hence, from the above,
We can conclude that
The solutions of the given equation are: -1 and 8

Question 18.
x + 4 = -x2
Answer:
The given equation is:
x + 4 = -x²
So,
x² + x + 4 = 0
The representation of the given equation in the coordinate plane is:

We know that,
The solutions of the equations are the x-intercepts of those equations when we draw the graph for them
Hence, from the above,
We can conclude that there are no real solutions for the given equation

Solve the equation using square roots.
Question 19.
4x2 = 64
Answer:
The given equation is:
4x² = 64
Now,

Hence, from the above,
We can conclude that the solutions of the given equation are: 4 and -4

Question 20.
-3x2 + 6 = 10
Answer:
The given equation is:
-3x² + 6 = 10

We know that,
The square of any number won’t be negative
Hence, from the above,
We can conclude that there are no real solutions for the given equation

Question 21.
(x – 8)2 = 1
Answer:
The given equation is:
(x – 8)² = 1
Now,

Hence, from the above,
We can conclude that the solutions of the given equation are: 7 and 9

Question 22.
Explain how to determine the number of real solutions of x2 = 100 without solving.
Answer:
The given equation is:
x² = 100
So,
x² – 100 = 0
We know that,
If the equation is in the form of
x² -a = 0
then, that equation has 2 real solutions
Hence, from the above,
We can conclude that the number of real solutions for the given equation is: 2

Question 23.
The length of a rectangular prism is four times its width. The volume of the prism is 380 cubic meters. Find the length and width of the prism.
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations q 4
Answer:
It is given that
The length of a rectangular prism is 4 times its width
The height of the rectangular prism is: 5m
So,
l = 4w
We know that,
The volume of the rectangular prism = l × w × h
So,
380 = w × 4w × 5
4w² = \(\frac{380}{5}\)
4w² = 76
w² = \(\frac{76}{4}\)
w² = 19
√w² = √19
w = √19 or w = -√19
We know that,
The width of the rectangular prism can’t be negative
So,
w = √19
So,
l = 4√19
Hence, from the above,
We can conclude that
The length of the rectangular prism is: 4√19 m
The width of the rectangular prism is: √19 m

Question 24.
You cast a fishing lure into the water from a height of 4 feet above the water. The height h (in feet) of the fishing lure after t seconds can be modeled by the equation h = -16t2 + 24t + 4.
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations q 5
a. After how many seconds does the fishing lure reach a height of 12 feet?
Answer:
It is given that the height (h) of the fishing lure after t seconds can be modeled by the equation
h = -16t² + 24t + 4
It is given that the height of the fishing lure is: 12 feet
So,
12 = -16t² + 24t + 4
-16t² + 24t = 12 – 4
-16t² + 24t = 8
-2t² + 3t = 1
2t² – 3t + 1 = 0
2t² – 2t – t + 1 = 0
2t (t – 1) – 1 (t – 1) = 0
(2t – 1) (t – 1) = 0
2t – 1 = 0 or t – 1 = 0
2t = 1 or t = 1
t = \(\frac{1}{2}\) or t = 1
Hence, from the above,
We can conclude that after around 1 second, the fishing lure will reach a height of 12 feet

b. After how many seconds does the fishing lure hit the water?
Answer:
From part (a),
The values of t are: [laex]\frac{1}{2}[/latex] and 1
So,
After 1 second, the fishing lure will hit the water

Lesson 9.4 Solving Quadratic Equations by Completing the Square

Essential Question How can you use “completing the square” to solve a quadratic equation?
Answer:
The steps to “Completing the Square” to solve a quadratic equation are:
Step 1:
Divide all terms by a (The coefficient of x²)
Step 2:
Move the number term (c / a) to the right side of the equation
Step 3:
Complete the square on the left side of the equation and balance this by adding the same value to the right side of the equation

EXPLORATION 1

Solving by Completing the Square
Work with a partner.
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 1
a. Write the equation modeled by the algebra tiles. This is the equation to be solved.
Answer:
The equation modeled by the algebra tiles as shown in the below figure is:

x² + 4x = -2

b. Four algebra tiles are added to the left side to “complete the square.” Why are four algebra tiles also added to the right side?
Answer:
The representation of the addition of four tiles on the left side is:

As shown in the above figure,
The four algebra tiles are added to the right side for the purpose of “Completing the square”
The same four algebra tiles also have to be added to the right side to balance the equation
So,
For the above purpose,
We added four algebra tiles to the right side also

c. Use algebra tiles to label the dimensions of the square on the left side and simplify on the right side.
Answer:

From the above figure,
Blue tile ——> Represents the coefficient of x²
Green tile —–> Represents the coefficient of x
Yellow and Red tiles —–> represent the constant terms

d. Write the equation modeled by the algebra tiles so that the left side is the square of a binomial. Solve the equation using square roots.
Answer:
From part (a),
The equation that can be modeled by the algebra tiles is:
x² + 4x = -2
Add with “4” on both sides as shown in part (b)
x² + 4x + 4 = -2 + 4
(x + 2)² = 2
√(x + 2)² = √2
x + 2 = √2 or x + 2 = -√2
x = √2 – 2 or x = -(√2 + 2)
Hence, from the above,
We can conclude that the values of x are: √2 – 2 and -(√2 + 2)

EXPLORATION 2

Solving by Completing the Square
Work with a partner.
a. Write the equation modeled by the algebra tiles.
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 2
Answer:
From the given algebra tiles representation,
The equation that can be modeled is:
x² + 6x = -5

b. Use algebra tiles to “complete the square.”
Answer:
From part (a),
The equation modeled by algebra tiles is:
x² + 6x = -5
To solve the quadratic equation by using algebra tiles,
Add 9 algebra tiles on both sides
So,
x² + 6x + 9 = -5 + 9
(x + 3)² = 4
Hence, from the above,
We can conclude that
The quadratic equation that is completed by using “Completing the squares” is:
(x + 3)² = 4

c. Write the solutions of the equation.
Answer:
From part (b),
The equation is:
(x + 3)² = 4
So,
√(x + 3)² = √4
x + 3 = 2 or x + 3 = -2
x = 2 – 3 or x = -2 – 3
x = -1 or x = -5
Hence, from the above,
We can conclude that the solutions of the given equation are: -1 and -5

d. Check each solution in the original equation.
Answer:
The original equation is:
(x + 3)² = 4
Now,
Put x = -1 and -5 in the above equation
For x = -1,
(-1 + 3)² = 4
(2)² = 4
4 = 4
For x = -5,
(-5 + 3)² = 4
(-2)² = 4
4 = 4

Communicate Your Answer

Question 3.
How can you use “completing the square” to solve a quadratic equation?
Answer:
The steps to “Completing the Square” to solve a quadratic equation are:
Step 1:
Divide all terms by a (The coefficient of x²)
Step 2:
Move the number term (c / a) to the right side of the equation
Step 3:
Complete the square on the left side of the equation and balance this by adding the same value to the right side of the equation

Question 4.
Solve each quadratic equation by completing the square.
a. x2 – 2x = 1
Answer:
The given equation is:
x² – 2x = 1
Step 1:
Divide all the terms with x² coefficient i.e., 1
So,
x² – 2x = 1
x² – 2x – 1 = 0
Step 2:
Move the constant term to the right side of the equation
So,
x² – 2x = 1
Step 3:
Add with 1 on both sides
So,
x² – 2x + 1 = 1 + 1
(x – 1)² = 2
Hence, from the above,
We can conclude that the solution of the given quadratic equation by completing the square is:
(x – 1)² = 2

b. x2 – 4x = -1
Answer:
The given equation is:
x² – 4x = -1
Step 1:
Divide all the terms with x² coefficient i.e., 1
So,
x² – 4x = -1
x² – 4x + 1 = 0
Step 2:
Move the constant term to the right side of the equation
So,
x² – 4x = -1
Step 3:
Add with 4 on both sides
So,
x² – 4x + 4 = -1 + 4
(x – 2)² = 3
Hence, from the above,
We can conclude that the solution of the given quadratic equation by completing the square is:
(x – 2)² = 3

c. x2 + 4x = -3
Answer:
The given equation is:
x² + 4x = -3
Step 1:
Divide all the terms with x² coefficient i.e., 1
So,
x² + 4x = -3
x² + 4x + 3 = 0
Step 2:
Move the constant term to the right side of the equation
So,
x² + 4x = -3
Step 3:
Add with 4 on both sides
So,
x² + 4x + 4 = -3 + 4
(x + 2)² = 1
Hence, from the above,
We can conclude that the solution of the given quadratic equation by completing the square is:
(x + 2)² = 1

Monitoring Progress

Complete the square for the expression. Then factor the trinomial.
Question 1.
x2 + 10x
Answer:
The given expression is:
x² + 10x
Compare the given expression with
ax² + bx
So,
a = 1 and b = 10
Now,
To complete the square of the following expression, the steps are:
Step 1:
Find the value of one-half of b
So,
\(\frac{b}{2}\) = \(\frac{10}{2}\)
\(\frac{b}{2}\) = 5
Step 2:
Square the result from step 1
So,
\(\frac{b²}{4}\) = 5² = 25
Step 3:
Add the result from step 2 to the given equation
So,
x² + 10x + 25 = (x + 5)²
Hence, from the above,
We can conclude that the factor of the given trinomial is: (x + 5)²

Question 2.
x2 – 4x
Answer:
The given expression is:
x² – 4x
Compare the given expression with
ax² + bx
So,
a = 1 and b = -4
Now,
To complete the square of the following expression, the steps are:
Step 1:
Find the value of one-half of b
So,
\(\frac{b}{2}\) = \(\frac{-4}{2}\)
\(\frac{b}{2}\) = -2
Step 2:
Square the result from step 1
So,
\(\frac{b²}{4}\) = (-2)² = 4
Step 3:
Add the result from step 2 to the given equation
So,
x² – 4x + 4 = (x – 2)²
Hence, from the above,
We can conclude that the factor of the given trinomial is: (x – 2)²

Question 3.
x2 + 7x
Answer:
The given expression is:
x² + 7x
Compare the given expression with
ax² + bx
So,
a = 1 and b = 7
Now,
To complete the square of the following expression, the steps are:
Step 1:
Find the value of one-half of b
So,
\(\frac{b}{2}\) = \(\frac{7}{2}\)
Step 2:
Square the result from step 1
So,
\(\frac{b²}{4}\)
= \(\frac{49}{4}\)
Step 3:
Add the result from step 2 to the given equation
So,
x² + 7x + \(\frac{49}{4}\) = (x + \(\frac{7}{2}\))²
Hence, from the above,
We can conclude that the factor of the given trinomial is: (x + \(\frac{7}{2}\))²

Solve the equation by completing the square. Round your solutions to the nearest hundredth, if necessary.
Question 4.
x2 – 2x = 3
Answer:
The given equation is:
x² – 2x = 3
Compare the given equation with
x² + bx = d
So,
b = -2 and d = 3
We know that,
To complete the trinomial by completing the square,
We have to add \(\frac{b²}{4}\)
So,
\(\frac{b²}{4}\) = \(\frac{4}{4}\) = 1
So,
Add 1 on both sides to the given equation
So,
x² – 2x + 1 = 3 + 1
(x – 1)² = 4
√(x – 1)² = √4
x – 1 = 2 or x – 1 = -2
x = 2 + 1 or x = -2 + 1
x = 3 or x = -1
Hence, from the above,
We can conclude that the solutions of the given equation are: 3 and -1

Question 5.
m2 + 12m = -8
Answer:
The given equation is:
m² + 12m = -8
Compare the given equation with
x² + bx = d
So,
b = 12 and d = -8
We know that,
To complete the trinomial by completing the square,
We have to add \(\frac{b²}{4}\)
So,
\(\frac{b²}{4}\) = \(\frac{144}{4}\) = 36
So,
Add 36 on both sides to the given equation
So,
m² + 12m + 36 = -8 + 36
(m + 6)² = 28
√(m + 6)² = √28
m + 6 = √28 or m + 6 = -√28
m = √28 – 6 or m = -(√28 + 6)
Hence, from the above,
We can conclude that the solutions of the given equation are: √28 – 6 and -(√28 + 6)

Question 6.
3g2 – 24g + 27 = 0
Answer:
The given equation is:
3g² – 24g + 27 = 0
Divide the given equation by 3 into both sides
So,
The given equation becomes
g² – 8g + 9 = 0
g² – 8g = -9
Now,
Compare the given equation with
x² + bx = d
So,
b = -8 and d = -9
We know that,
To complete the trinomial by completing the square,
We have to add \(\frac{b²}{4}\)
So,
\(\frac{b²}{4}\) = \(\frac{64}{4}\) = 16
So,
Add 16 on both sides to the given equation
So,
g² – 8g + 16 = -9 + 16
(g – 4)² = 7
√(g – 4)² = √7
g – 4 = √7 or g – 4 = -√7
g = √7 + 4 or g = -√7 + 4
Hence, from the above,
We can conclude that the solutions of the given equation are: √7 + 4 and -√7 + 4

Determine whether the quadratic function has a maximum or minimum value. Then find the value.
Question 7.
y = -x2 – 4x + 4
Answer:
The given equation is:
y = -x² – 4x + 4
Subtract with 4 on both sides
y – 4 = -x²  – 4x + 4 – 4
y – 4 = -x²  – 4x
y – 4 = – (x² + 4x)
To complete the square of the expression present in the right side,
Subtract with 4 on both sides
So,
y – 4 – 4 = -(x² + 4x + 4)
y  – 4= -(x + 2)²
y = -(x + 2)² + 4

From the graph,
We can observe that the graph is in a closed shape
So,
The given equation will have a minimum value
Hence, from the above,
We can conclude that the minimum value of the given equation is: -4

Question 8.
y = x2 + 12x + 40
Answer:
The given equation is:
y = x² + 12x + 40
Subtract with 40 on both sides
y – 40 = x²  + 12x + 40 – 40
y – 40 = x²  + 12x
To complete the square of the expression present in the right side,
Add with 36 on both sides
So,
y – 40 + 36 = x² + 12x + 36
y  – 4= (x + 6)²
y = (x + 6)² + 4

From the graph,
We can observe that the graph is in an open shape
So,
The given equation will have a maximum value
Hence, from the above,
We can conclude that the maximum value of the given equation is: 4

Question 9.
y = x2 – 2x – 2
Answer:
The given equation is:
y = x² – 2x – 2
Add with 2 on both sides
y + 2 = x²  – 2x – 2 + 2
y + 2 = x²  – 2x
To complete the square of the expression present in the right side,
Add with 1 on both sides
So,
y + 2 + 1 = x² – 2x + 1
y  + 3= (x – 1)²
y = (x – 1)² – 3

From the graph,
We can observe that the graph is in an open shape
So,
The given equation will have a maximum value
Hence, from the above,
We can conclude that the maximum value of the given equation is: 3

Determine whether the function could be represented by the graph in Example 6. Explain.
Question 10.
h(x) = (x – 8)2 + 10
Answer:
The given equation is:
h (x) = (x – 8)² + 10
So,
From the above equation,
The vertex point that is the solution for the given equation is: (8, 10)
Since the values of x and y are positive, the graph must be in the 1st quadrant
So,
The representation of the given equation in the coordinate plane is:

Hence, from the above,
We can conclude that the given function represented by the graph

Question 11.
n(x) = -2(x – 5)(x – 20)
Answer:
The given equation is:
n (x) = -2 (x – 5) (x – 20)
So,
From the above equation,
The vertex point that is the solution for the given equation is: (-5, -20)
Since the values of x and y are positive, the graph must be in the 3rd quadrant
So,
The representation of the given equation in the coordinate plane is:

Hence, from the above,
We can conclude that the given function is not represented by the graph

Question 12.
WHAT IF?
Repeat Example 7 when the function is y = -16x2 + 128x.
Answer:
The given function is:
y = -16x² + 128x
y = -16(x² – 8x)
Now,
We have to solve the expression on the left side by using the completing the squares
So,
Subtract with 16 on both sides
y – 16 = -16(x² – 8x + 16)
y – 16 = -16(x – 4)²
y = -16 (x – 4)² + 16
a) The maximum height of the rocket:
From the above equation,
We can observe that the maximum value is: 16
Hence, from the above,
We can conclude that the model rocket reaches a maximum height of 16 feet
b) The axes of symmetry:
From the given equation that we obtained in the vertex form,
The vertex is: (4, 16)
Hence, from the above,
We can conclude that
The axis of symmetry is:
x = 4

Question 13.
WHAT IF?
You want the chalkboard to cover 4 square feet. Find the width of the border to the nearest inch.
Answer:
From Example 8,

From the above figure,
We can observe that the chalkboard is in the form of a rectangle
So,
From the above figure,
The length of the chalkboard = 7 – 2x
The width of the chalkboard = 3 – 2x
We know that,
Area of the rectangle = Length × Width
So,
4 = (7 – 2x) (3 – 2x)
21 – 14x – 6x + 4x² = 4
4x² – 20x + 21 = 4
4x² – 20x = 4 – 21
4x² – 20x = -17
4 (x² – 5x) = -17
x² – 5x = –\(\frac{17}{4}\)
To complete the square of the expression present on the left side,
Add with \(\frac{25}{4}\) on both sides
So,
x² – 5x + \(\frac{25}{4}\) = –\(\frac{17}{4}\) + \(\frac{25}{4}\)
x² – 5x + \(\frac{25}{4}\) = 2
(x – \(\frac{5}{2}\))² = 2
√(x – \(\frac{5}{2}\))² = √2
x – \(\frac{5}{2}\) = √2 or x – \(\frac{5}{2}\) = -√2
We know that,
√2 = 1.414
\(\frac{5}{2}\) = 2.5
So,
x = 1.414 + 2.5 or x = -1.414 + 2.5
x = 3.91 or x = 1.08
Since the width of the border is 3 feet,
x = 3.91 is not possible
So,
The width of the border = 1.08 foot
To convert the foot in to inches,
Multiply 1.08 with 12
So,
1.08 × 12 inches = 12.96 inches
Hence, from the above,
We can conclude that the width of the border in inches is: 12.96 inches

Solving Quadratic Equations by Completing the Square 9.4 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
The process of adding a constant c to the expression x2 + bx so that x2 + bx + c is a perfect square trinomial is called ________________.
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 1

Question 2.
VOCABULARY
Explain how to complete the square for an expression of the form x2 + bx.
Answer:
The steps that required to complete the square for an expression x² + bx are:
Step 1:
Find one half of b i.e., \(\frac{b}{2}\)
Step 2:
Find \(\frac{b²}{4}\) and add it to the expression x² + bx
So,
The equation will become
x² + bx + \(\frac{b²}{4}\)

Question 3.
WRITING
Is it more convenient to complete the square for x2 + bx when b is odd or when b is even? Explain.
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 3

Question 4.
WRITING
Describe how you can use the process of completing the square to find the maximum or minimum value of a quadratic function.
Answer:
There are two ways to find the absolute maximum/minimum value for f(x) = ax² + bx + c
Put the quadratic in standard form
f(x) = a(x − h)² + k, and the absolute maximum/minimum value is k and it occurs at x = h.
If a > 0, then the parabola opens up, and it is a minimum functional value of f (x)

Monitoring Progress and Modeling with Mathematics

In Exercises 5–10, find the value of c that completes the square.
Question 5.
x2 – 8x + c
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 5

Question 6.
x2 – 2x + c
Answer:
The given expression is:
x² – 2x + c
Compare the given expression with x² + bx + c
So,
b = -2
We know that,
The representation of the expression that is completed by using completing the squares is:
x² + bx + \(\frac{b²}{4}\)
So,
c = \(\frac{b²}{4}\)
c = \(\frac{4}{4}\)
c = 1
Hence, from the above,
We can conclude that the value of c is: 1

Question 7.
x2 + 4x + c
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 7

Question 8.
x2 + 12x + c
Answer:
The given expression is:
x² + 12x + c
Compare the given expression with x² + bx + c
So,
b = 12
We know that,
The representation of the expression that is completed by using completing the squares is:
x² + bx + \(\frac{b²}{4}\)
So,
c = \(\frac{b²}{4}\)
c = \(\frac{144}{4}\)
c = 36
Hence, from the above,
We can conclude that the value of c is: 36

Question 9.
x2 – 15x + c
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 9

Question 10.
x2 + 9x + c
Answer:
The given expression is:
x² + 9x + c
Compare the given expression with x² + bx + c
So,
b = 9
We know that,
The representation of the expression that is completed by using completing the squares is:
x² + bx + \(\frac{b²}{4}\)
So,
c = \(\frac{b²}{4}\)
c = \(\frac{81}{4}\)
Hence, from the above,
We can conclude that the value of c is: \(\frac{81}{4}\)

In Exercises 11–16, complete the square for the expression. Then factor the trinomial.
Question 11.
x2 – 10x
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 11

Question 12.
x2 – 40x
Answer:
The given expression is:
x² – 40x
Compare the given expression with
ax² + bx
So,
a = 1 and b = -40
Now,
To complete the square of the following expression, the steps are:
Step 1:
Find the value of one-half of b
So,
\(\frac{b}{2}\) = \(\frac{-40}{2}\)
\(\frac{b}{2}\) = -20
Step 2:
Square the result from step 1
So,
\(\frac{b²}{4}\) = (-20)² = 400
Step 3:
Add the result from step 2 to the given equation
So,
x² – 40x + 200 = (x – 20)²
Hence, from the above,
We can conclude that the factor of the given trinomial is: (x – 20)²

Question 13.
x2 + 16x
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 13

Question 14.
x2 + 22x
Answer:
The given expression is:
x² + 22x
Compare the given expression with
ax² + bx
So,
a = 1 and b = 22
Now,
To complete the square of the following expression, the steps are:
Step 1:
Find the value of one-half of b
So,
\(\frac{b}{2}\) = \(\frac{22}{2}\)
\(\frac{b}{2}\) = 11
Step 2:
Square the result from step 1
So,
\(\frac{b²}{4}\) = 11² = 121
Step 3:
Add the result from step 2 to the given equation
So,
x² + 22x + 121 = (x + 11)²
Hence, from the above,
We can conclude that the factor of the given trinomial is: (x + 11)²

Question 15.
x2 + 5x
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 15

Question 16.
x2 – 3x
Answer:
The given expression is:
x² – 3x
Compare the given expression with
ax² + bx
So,
a = 1 and b = -3
Now,
To complete the square of the following expression, the steps are:
Step 1:
Find the value of one-half of b
So,
\(\frac{b}{2}\) = \(\frac{-3}{2}\)
\(\frac{b}{2}\) = –\(\frac{3}{2}\)
Step 2:
Square the result from step 1
So,
\(\frac{b²}{4}\) = \(\frac{9}{4}\)
Step 3:
Add the result from step 2 to the given equation
So,
x² – 3x + \(\frac{9}{4}\) = (x – \(\frac{3}{2}\))²
Hence, from the above,
We can conclude that the factor of the given trinomial is: (x – \(\frac{3}{2}\))²

In Exercises 17–22, solve the equation by completing the square. Round your solutions to the nearest hundredth, if necessary.
Question 17.
x2 + 14x = 15
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 17

Question 18.
x2 – 6x = 16
Answer:
The given equation is:
x² – 6x = 16
To solve the given equation by using the completing the squares
Add 9 on both sides
So,
x² – 6x + 9 = 16 + 9
(x – 3)² = 25
√(x – 3)² = √25
x – 3 = 5 or x – 3 = -5
x = 5 + 3 or x = -5 + 3
x = 8 or x = -2
Hence, from the above,
We can conclude that the values of x are: 8 and -2

Question 19.
x2 – 4x = -2
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 19

Question 20.
x2 + 2x = 5
Answer:
The given equation is:
x² + 2x = 5
To solve the given equation by using the completing the squares
Add 1 on both sides
So,
x² + 2x + 1 = 5 + 1
(x + 1)² = 6
√(x + 1)² = √6
x + 1 = √6 or x + 1 = -√6
We know that,
√6 = 2.44
So,
x = 2.44 – 1 or x = -2.44 – 1
x = 1.44 or x = -3.44
Hence, from the above,
We can conclude that the values of x are: 1.44 and -3.44

Question 21.
x2 – 5x = 8
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 21

Question 22.
x2 + 11x = -10
Answer:
The given equation is:
x² + 11x = -10
To solve the given equation by using the completing the squares,
Add with \(\frac{121}{4}\) on both sides
So,
x² + 11x + \(\frac{121}{4}\) = -10 + \(\frac{121}{4}\)
(x + 11)² = \(\frac{81}{4}\)
√(x + 11)² = √\(\frac{81}{4}\)
x + 11 = √\(\frac{81}{4}\) or x + 11 = -√\(\frac{81}{4}\)
We know that,
\(\frac{81}{4}\) = 20.25
So,
x = 20.25 – 11 or x = -20.25 – 11
x = 9.25 or x = -31.25
Hence, from the above,
We can conclude that the values of x are: 9.25 and -31.25

Question 23.
MODELING WITH MATHEMATICS
The area of the patio is 216 square feet.
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 3
a. Write an equation that represents the area of the patio.
b. Find the dimensions of the patio by completing the square.
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 23

Question 24.
MODELING WITH MATHEMATICS
Some sand art contains sand and water sealed in a glass case, similar to the one shown. When the art is turned upside down, the sand, and waterfall to create a new picture. The glass case has a depth of 1 centimeter and a volume of 768 cubic centimeters.
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 4
a. Write an equation that represents the volume of the glass case.
Answer:
It is given that the glass case has a depth of 1 centimeter and a volume of 768 cubic centimeters
We know that,
The volume of the glass case = L × W × H
So,
768 = x (x – 8) × 1
x² – 8x = 768
Hence, from the above,
We can conclude that the equation that represents the volume of the glass case is:
x² – 8x = 768

b. Find the dimensions of the glass case by completing the square.
Answer:
From part (a),
The equation that represents the volume of the glass case is:
x² – 8x = 768
To solve the above equation by using the completing the squares,
Add 16 on both sides
So,
x² – 8x + 16 = 768 + 16
(x – 4)² = 784
√(x – 4)² = √784
x – 4 = 28 or x – 4 = -28
x = 28 + 4 or x = -28 + 4
x = 32 or x = -24
We know that,
The length and width won’t be negative
So,
x = 32
So,
Length of the glass case = x – 8 = 32 – 8 =24 cm
Width of the glass case = x = 32 cm
Hence, from the above,
We can conclude that
The length of the glass case is: 24 cm
The width of the glass case is: 32 cm

In Exercises 25–32, solve the equation by completing the square. Round your solutions to the nearest hundredth, if necessary.
Question 25.
x2 – 8x + 15 = 0
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 25

Question 26.
x2 + 4x – 21 = 0
Answer:
The given equation is:
x² + 4x – 21 = 0
Add with 21 on both sides
So,
x² + 4x – 21 + 21 = 0 + 21
x² + 4x = 21
To solve the given equation by using the completing the squares,
Add 4 on both sides
So,
x² + 4x + 4 = 21 + 4
(x + 2)² = 25
√(x + 2)² = √25
x + 2 = 5 or x + 2 = -5
x = 5 – 2 or x = -5 – 2
x = 3 or x = -7
Hence, from the above,
We can conclude that
The values of x for the given equation are: 3 and -7

Question 27.
2x2 + 20x + 44 = 0
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 27

Question 28.
3x2 – 18x + 12 = 0
Answer:
The given equation is:
3x² – 18x + 12 = 0
3 (x² – 6x + 4) = 0
x² – 6x + 4 = 0
Subtract with 4 on both sides
So,
x² – 6x – 4 + 4 = 0 + 4
x² – 6x = 4
To solve the given equation by using the completing the squares,
Add 9 on both sides
So,
x² – 6x + 9 = 9 + 4
(x – 3)² = 13
√(x – 3)² = √13
We know that,
√13 = 3.60
So,
x – 3 = 3.60 or x – 3 = -3.60
x = 3.60 + 3 or x = -3.60 + 3
x = 6.60 or x = -0.60
Hence, from the above,
We can conclude that
The values of x for the given equation are: 6.60 and -0.60

Question 29.
-3x2 – 24x + 17 = -40
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 29

Question 30.
-5x2 – 20x + 35 = 30
Answer:
The given equation is:
-5x² – 20x + 35 = 30

We know that,
√5 = 2.23
So,
x = -2 – 2.23 or x = -2 + 2.23
x = -5.23 or x = 0.23
Hence, from the above,
We can conclude that
The values of x for the given equation are: 0.23 and -5.23

Question 31.
2x2 – 14x + 10 = 26
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 31

Question 32.
4x2 + 12x – 15 = 5
Answer:
The given equation is:
4x² + 12x – 15 = 5

We know that,
√29 = 5.38
So,
x = 1.19 or x = -6.88
Hence, from the above,
We can conclude that
The values of x for the given equation are: -6.88 and 1.19

Question 33.
ERROR ANALYSIS
Describe and correct the error in solving x2 + 8x = 10 by completing the square.
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 5
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 33

Question 34.
ERROR ANALYSIS
Describe and correct the error in the first two steps of solving 2x2 – 2x – 4 = 0 by completing the square.
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 6
Answer:
The given equation is:
2x² – 2x – 4 = 0
Compare the given equation with
ax² + bx + c = 0
To solve the given equation by using the completing the squares,
The following steps are:
Step 1:
Divide the given equation with the coefficient of x² i.e., 2
So,
x² – x – 2 = 0
Step 2:
Move the term (c / a) to the right side of the equation
So,
x² – x = 2
Step 3:
Complete the  equation by using the completing the squares
So,
Add \(\frac{1}{4}\) on both sides
x² – x + \(\frac{1}{4}\) = 2+ \(\frac{1}{4}\)
Hence,
(x – \(\frac{1}{2}\))² = \(\frac{9}{4}\)

Question 35.
NUMBER SENSE
Find all values of b for which x2 + bx + 25 is a perfect square trinomial. Explain how you found your answer.
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 35

Question 36.
REASONING
You are completing the square to solve 3x2 + 6x = 12. What is the first step?
Answer:
The given equation is:
3x² + 6x = 12
To solve the given equation by using the completing the square,
The first step we have to follow is:
Divide the given equation with the coefficient of x²
Hence,
x² + 2x = 4

In Exercises 37–40, write the function in vertex form bycompleting the square. Then match the function with its graph.
Question 37.
y = x2 + 6x + 3
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 37

Question 38.
y = -x2 + 8x – 12
Answer:
The given equation is:
y = -x² + 8x – 12
Add with 12 on both sides
So,
y + 12 = -x² + 8x – 12+ 12
y + 12 = -x² + 8x
y + 12 = – (x² – 8x)
To solve the given equation by using the completing the squares,
Subtract with 16 on both sides
So,
y + 12 – 16 = – (x² – 8x + 16)
y – 4 = – (x – 4)²
y = -(x – 4)² + 4
So,
From the vertex form of the equation,
The vertex point is: (-4, 4)
Hence, from the above,
We can conclude that
The vertex point of the given equation is: (-4, 4)
The vertex point matches with graph A

Question 39.
y = -x2 – 4x – 2
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 39

Question 40.
y = x2 – 2x + 4
Answer:
The given equation is:
y = x² – 2x + 4
Subtract with 4 on both sides
y – 4 = x² – 2x + 4 – 4
y – 4 = x² – 2x
To solve the given equation by using the completing the squares,
We have to add 1 on both sides
So,
y – 4 + 1 = x² – 2x + 1
y – 3 = (x – 1)²
y = (x – 1)² + 3
So,
From the vertex form of the equation,
The vertex point is: (1, 3)
Hence, from the above,
We can conclude that
The vertex point of the given equation is: (1, 3)
The vertex point matches with graph C

Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 7

In Exercises 41–46, determine whether the quadratic function has a maximum or minimum value. Then find the value.
Question 41.
y = x2 – 4x – 2
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 41

Question 42.
y = x2 + 6x + 10
Answer:
The given equation is:
y = x² + 6x + 10
Subtract with 10 on both sides
y – 10 = x² + 6x + 10 – 10
y – 10 = x² + 6x
To solve the given equation by using the completing the squares,
We have to add 9 on both sides
So,
y – 10 + 9 = x² + 6x + 9
y – 1 = (x + 3)²
y = (x + 3)² + 1
So,
From the vertex form of the equation,
The vertex point is: (-3, 1)
From the above vertex point,
We can observe that the x-coordinate is negative
So,
The parabola closes down and the y-coordinate will be the maximum value of the given equation
Hence, from the above,
We can conclude that
The given equation has a maximum value and the maximum value of the given equation is: 1

Question 43.
y = -x2 – 10x – 30
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 43

Question 44.
y = -x2 + 14x – 34
Answer:
The givne equation is:
y = -x² + 14x – 34
Add with 34 on both sides
y + 34 = -x² + 14x + 34 – 34
y + 34 = -x² + 14x
y + 34 = – (x² – 14x)
To solve the given equation by using the completing the squares,
We have to subtract 49 on both sides
So,
y + 34 – 49 = – (x² – 14x + 49)
y – 15 = -(x – 7)²
y = -(x – 7)² + 15
So,
From the vertex form of the equation,
The vertex point is: (-7, 15)
From the above vertex point,
We can observe that the x-coordinate is negative
So,
The parabola closes down and the y-coordinate will be the maximum value of the given equation
Hence, from the above,
We can conclude that
The given equation has a maximum value and the maximum value of the given equation is: 15

Question 45.
f(x) = -3x2 – 6x – 9
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 45

Question 46.
f(x) = 4x2 – 28x + 32
Answer:
The given equation is:
f (x) = 4x² – 28x + 32
Subtract with 32 on both sides
f (x) – 32 = 4x² – 28x + 32 – 32
f (x) – 32 = 4x² – 28x
f (x) – 32 = 4 (x² – 7x)
To solve the given equation by using the completing the squares,
We have to add \(\frac{49}{4}\) on both sides
So,
f (x) – 32 + \(\frac{49}{4}\) =4 ( x² – 7x + \(\frac{49}{4}\))
y – \(\frac{51}{4}\) = 4 (x – \(\frac{7}{2}\))²
y = 4(x – \(\frac{7}{2}\))² + \(\frac{51}{4}\)
So,
From the vertex form of the equation,
The vertex point is: (14, \(\frac{51}{4}\))
From the above vertex point,
We can observe that the x-coordinate is positive
So,
The parabola opens up and the y-coordinate will be the minimum value of the given equation
Hence, from the above,
We can conclude that
The given equation has a minimum value and the minimum value of the given equation is: \(\frac{51}{4}\)

In Exercises 47–50, determine whether the graph could represent the function. Explain.
Question 47.
y = -(x + 8)(x + 3)
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 8
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 47

Question 48.
y = (x – 5)2
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 9
Answer:
The given equation is:
y = (x – 5)²
y = (x – 5) (x – 5)
So,
The vertex point for the given equation is: (5, 5)
Now,
From the given vertex point,
We can observe that the parabola closes up since the x-coordinate is positive and we will have a minimum value
The minimum value will be the y-coordinate of the equation in the vertex form
Hence, from the above,
We can conclude that the given graph matches with the given equation

Question 49.
y = \(\frac{1}{4}\)(x + 2)2 – 4
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 10
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 49

Question 50.
y = -2(x – 1)(x + 2)
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 11
Answer:
The given equation is:
y = -2 (x – 1) (x + 2)
So,
The vertex point for the given equation is: (-2, 2)
Now,
From the given vertex point,
We can observe that the parabola opens down since the x-coordinate is negative and we will have a maximum value
The maximum value will be the y-coordinate of the equation in the vertex form
Hence, from the above,
We can conclude that the given graph matches with the given equation

In Exercises 51 and 52, determine which of the functions could be represented by the graph. Explain.
Question 51.
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 12
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 51.1
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 51.2

Question 52.
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 13
Answer:

Question 53.
MODELING WITH MATHEMATICS
The function h = -16t2 + 48t represents the height h (in feet) of a kickball t seconds after it is kicked from the ground.
a. Find the maximum height of the kickball.
b. Find and interpret the axis of symmetry.
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 53

Question 54.
MODELING WITH MATHEMATICS
You throw a stone from a height of 16 feet with an initial vertical velocity of 32 feet per second. The function h = -16t2 + 32t + 16 represents the height h (in feet) of the stone after t seconds.
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 14
a. Find the maximum height of the stone.
Answer:
We know that,
The maximum height of the stone is:
Hmax = h + \(\frac{v²}{2g}\)
We know that,
g = 10 m/s
So,
Hmax = 16 + \(\frac{1024}{20}\)
Hmax = 67.2 feet

b. Find and interpret the axis of symmetry.
Answer:
The given equation is:
h = -16t² + 32t + 16
Subtract with 16  on both sides
h – 16 = -16t² + 32t + 16 – 16
h – 16 = -16t² + 32t
h – 16 = -16 (t² – 2t)
To solve the given equation by completing the squares,
h – 16 – 16 = -16 (t² – 2t + 1)
h – 32 = -16 (t – 1)²
h = -16 (t – 1)² + 32
The above equation is in the form of the vertex
So,
The axes of symmetry will be the t-axis
Hence, from the above,
We can conclude that the axis of symmetry for the given equation is:
t = 1

Question 55.
MODELING WITH MATHEMATICS
You are building a rectangular brick patio surrounded by a crushed stone border with a uniform width, as shown. You purchase patio bricks to cover 140 square feet. Find the width of the border.
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 15
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 55.1
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 55.2

Question 56.
MODELING WITH MATHEMATICS
You are making a poster that will have a uniform border, as shown. The total area of the poster is 722 square inches. Find the width of the border to the nearest inch.
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 16
Answer:
It is given that the total area of the poster is: 722 square inches
From the given figure,
The width of the poster is: (28 – 2x) inches
The length of the poster is: (22 – 2x) inches
We know that,
The area of the poster = Length × Width
722 = (28 – 2x) (22 – 2x)
722 = 616 – 56x – 44x + 4x²
722 – 616 = 4x² – 56x – 44x
106 = 4x² – 100x
Divide by 4 on both sides
So,
\(\frac{53}{2}\) = x² – 25x
To solve the above equation by using the completing the squares,
Add with \(\frac{625}{4}\) on both sides
So,
\(\frac{53}{2}\) + \(\frac{625}{4}\) = x² – 25x + \(\frac{625}{4}\)
\(\frac{731}{4}\) = (x – \(\frac{25}{2}\))²
√(x – \(\frac{25}{2}\))² = √\(\frac{731}{4}\)
x – \(\frac{25}{2}\) = √\(\frac{731}{4}\) or x – \(\frac{25}{2}\) = -√\(\frac{731}{4}\)
We know that,
√\(\frac{731}{4}\) = 13.62
\(\frac{25}{2}\) = 12.5
So,
x = 13.62 + 12.5 or x = -13.62 + 12.5
x = 26.12 or x = -1.12
From the given figure,
We can observe that the width of the poster is 28 inches and the width of the poster won’t be negative
So,
The width of the border is: -1.12 inches
So,
The width of the border = 28 – 2x
= 28 – 2 (-1.12)
= 30.24 inches
Hence, from the above,
We can conclude that the width of the border is: 30.24 inches

MATHEMATICAL CONNECTIONS In Exercises 57 and 58, find the value of x. Round your answer to the nearest hundredth, if necessary.
Question 57.
A = 108 m2
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 17
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 57

Question 58.
A = 288 in.2
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 18
Answer:
The given figure is:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 18
From the given figure,
We can say that it is a rectangle
We know that,
The area of the rectangle = Length × Width
So,
288 = (2x + 10) (3x)
288 = 6x² + 30x
Divide by 6 into both sides
So,
48 = x² + 5x
To solve the above equation by using the completing the squares,
Add with \(\frac{25}{4}\) on both sides
So,
48 + \(\frac{25}{4}[latex] = x² + 5x + [latex]\frac{25}{4}\)
54.25 = (x + \(\frac{5}{2}\))²
√(x + \(\frac{5}{2}\))² = √54.25
We know that,
√54.25 = 7.36
\(\frac{5}{2}\) = 2.5
So,
x + \(\frac{5}{2}\) = 7.36 or x + \(\frac{5}{2}\) = -7.36
x = 7.36 – 2.5 or x = -7.36 – 2.5
x = 4.86 or x = -9.86
We know that,
The length and width won’t be negative
So,
x = 4.86
Hence, from the above,
We can conclude that the value of x is: 4.86

In Exercises 59–62, solve the equation by completing the square. Round your solutions to the nearest hundredth, if necessary.
Question 59.
0.5x2 + x – 2 = 0
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 59

Question 60.
0.75x2 + 1.5x = 4
Answer:
The given equation is:
0.75x² + 1.5x = 4

Hence, from the above,
We can conclude that the values of x are: 1.51 and -3.51

Question 61.
\(\frac{8}{3}\) x – \(\frac{2}{3}\)x2 = –\(\frac{5}{6}\)
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 61

Question 62.
\(\frac{1}{4}\)x2 + \(\frac{1}{2}\)x – \(\frac{5}{4}\) = 0
Answer:
The given equation is:
\(\frac{1}{4}\)x2 + \(\frac{1}{2}\)x – \(\frac{5}{4}\) = 0

Hence, from the above,
We can conclude that the values of x are: 1.44 and -3.44

Question 63.
PROBLEM-SOLVING
The distance d (in feet) that it takes a car to come to a complete stop can be modeled by d = 0.05s2 + 2.2s, where s is the speed of the car (in miles per hour). A car has 168 feet to come to a complete stop. Find the maximum speed at which the car can travel.
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 63

Question 64.
PROBLEM SOLVING
During a “big air” competition, snowboarders launch themselves from a half-pipe, perform tricks in the air, and land back in the half-pipe. The height h (in feet) of a snowboarder above the bottom of the half-pipe can be modeled by h = -16t2 + 24t + 16.4, where t is the time (in seconds) after the snowboarder launches into the air. The snowboarder lands 3.2 feet lower than the height of the launch. How long is the snowboarder in the air? Round your answer to the nearest tenth of a second.
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 19
Answer:

Question 65.
PROBLEM SOLVING
You have 80 feet of fencing to make a rectangular horse pasture that covers 750 square feet. A barn will be used as one side of the pasture, as shown.
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 20
a. Write equations for the amount of fencing to be used and the area enclosed by the fencing.
b. Use substitution to solve the system of equations from part (a). What are the possible dimensions of the pasture?
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 65.1
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 65.2

Question 66.
HOW DO YOU SEE IT?
The graph represents the quadratic function y = x2 – 4x + 6.
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 21
a. Use the graph to estimate the x-values for which y = 3.
Answer:
The given quadratic equation is:
y = x² – 4x + 6
It is given that the value of y is: 3
So,
x² – 4x + 6 = 3
x² – 4x = 3 – 6
x² – 4x = -3
To solve the equation by using the completing the squares,
Add with 4 on both sides
So,
x² – 4x + 4 = -3 + 4
(x – 2)² = 1
√(x – 2)² = √1
x – 2 = 1 or x – 2 = -1
x = 1 + 2 or x = -1 + 2
x = 3 or x = 1
Hence, from the above,
We can conclude that,
For y = 3, the values of x are: 3 and 1

b. Explain how you can use the method of completing the square to check your estimates in part (a).
Answer:
The steps to “Completing the Square” to check the estimates in part (a) are:
Step 1:
Divide all terms by a (The coefficient of x²)
Step 2:
Move the number term (c / a) to the right side of the equation
Step 3:
Complete the square on the left side of the equation and balance this by adding the same value to the right side of the equation

Question 67.
COMPARING METHODS
Consider the quadratic equation x2 + 12x + 2 = 12.
a. Solve the equation by graphing.
b. Solve the equation by completing the square.
c. Compare the two methods. Which do you prefer? Explain.
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 67.1
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 67.2

Question 68.
THOUGHT-PROVOKING
Sketch the graph of the equation x2 – 2xy + y2 – x – y = 0. Identify the graph.
Answer:
The given equation is:
x² – 2xy + y² – x – y = 0
Compare the given equation with the equation of the circle
x² + y² + 2gx + 2fy + 2hxy + c = 0
So,
By comparing the two equations,
We can observe that the 2 equations are the same
Hence, from the above,
We can conclude that the graph of the given equation looks like a circle

Question 69.
REASONING
The product of two consecutive even integers that are positive is 48. Write and solve an equation to find the integers.
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 69

Question 70.
REASONING
The product of two consecutive odd integers that are negative is 195. Write and solve an equation to find the integers.
Answer:
Let the 2 consecutive odd integers be 1 – n and 3 – n
It is given that the product of two consecutive negative odd integers si 195
So,
(1 – n) (3 – n) = 195

Hence, from the above
We can conclude that the 2 integers are: -12 and 16

Question 71.
MAKING AN ARGUMENT
You purchase stock for $16 per share. You sell the stock 30 days later for $23.50 per share. The price y (in dollars) of a share during the 30-day period can be modeled by y = -0.025x2 + x + 16, where x is the number of days after the stock is purchased. Your friend says you could have sold the stock earlier for $23.50 per share. Is your friend correct? Explain.
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 22
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 71

Question 72.
REASONING
You are solving the equation x2 + 9x =18. What are the advantages of solving the equation by completing the square instead of using other methods you have learned?
Answer:
The given equation is:
x² + 9x = 18
x² + 2( \(\frac{9}{2}\) )x = 18
Now,
From the above equation,
We can observe that
To make the expression present on the left side a factor trinomial to get the exact values of x, we have to use the “Completing the squares” method
When we use the graphing method,
We won’t get the exact values of x for the equation
Hence, from the above,
We can conclude that we have to solve the above equation by using the “Completing the squares” method to get the exact values of x when we compared with the other methods

Question 73.
PROBLEM-SOLVING
You are knitting a rectangular scarf. The pattern results in a scarf that is 60 inches long and 4 inches wide. However, you have enough yarn to knit 396 square inches. You decide to increase the dimensions of the scarf so that you will use all your yarn. The increase in the length is three times the increase in the width. What are the dimensions of your scarf?
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 23
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 73.1
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 73.2

Question 74.
WRITING
How many solutions does x2 + bx = c have when c < -(\(\frac{b}{2}\))2 ? Explain.
Answer:
The given equation is:
x² + bx = c
x² + bx + \(\frac{b²}{4}\) = c + \(\frac{b²}{4}\)
(x + \(\frac{b}{2}\))² = c +c + \(\frac{b²}{4}\)
It is given that
c < – \(\frac{b²}{4}\)
We know that,
When c < – \(\frac{b²}{4}\), the square value will also be negative
But, we know that the square won’t take any negative values
Hence, from the above,
We can conclude that the number of solutions of x² + bx = c when c <- \(\frac{b²}{4}\) are: 0

Maintaining Mathematical Proficiency

Write a recursive rule for the sequence.
Question 75.
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 24
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 75

Question 76.
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 25
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 25
From the given graph,
We can observe that
The first term is: 3 (When x = 1)
Each term is obtained by multiplying the previous term by 2
Hence,
The recursive rule for the sequence is:
a1 = an-1 × 2

Question 77.
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 26
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 77

Simplify the expression \(\sqrt{b^{2}-4 a c}\) for the given values.
Question 78.
a = 3, b = -6, c = 2
Answer:
The given expression is:
\(\sqrt{b² – 4ac}\)
It is given that
a = 3, b = -6, and c = 2
So,
\(\sqrt{b² – 4ac}\)
= \(\sqrt{(-6)² – 4 (3) (2)}\)
= \(\sqrt{36 – 24}\)
= \(\sqrt{12}\)
= \(\sqrt{4 × 3}\)
= 2\(\sqrt{3}\)

Question 79.
a = -2, b = 4, c = 7
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 9 Solving Quadratic Equations 9.4 a 79

Question 80.
a = 1, b = 6, c = 4
Answer:
The given expression is:
\(\sqrt{b² – 4ac}\)
It is given that
a = 1, b = 6, and c = 4
So,
\(\sqrt{b² – 4ac}\)
= \(\sqrt{(6)² – 4 (1) (4)}\)
= \(\sqrt{36 – 16}\)
= \(\sqrt{20}\)
= \(\sqrt{4 × 5}\)
= 2\(\sqrt{5}\)

Lesson 9.5 Solving Quadratic Equations Using the Quadratic Formula

Essential Question How can you derive a formula that can be used to write the solutions of any quadratic equation in standard form?
Answer:
The formula can be actually derived using the steps involved in completing the square.
It stems from the fact that any quadratic function or equation of the form
y = a x² + b x + c
can be solved for its roots.

EXPLORATION 1

Deriving the Quadratic Formula
Work with a partner. The following steps show a method of solving ax2 + bx + c = 0. Explain what was done in each step.
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 1
Answer:
Step 1:
Write the equation
Step 2:
Multiply with “4a” on both sides
Step 3:
Add b² on both sides
Step 4:
Subtract with “4ac” on both sides
Step 5:
Write the expression that is present on the left side i.e., in the form of a trinomial i.e., (x ± a)²
Step 6:
Take “Square root” on both sides
Step 7:
Subtract with “b” on both sides
Step 8:
Divide by “2a” into both sides

EXPLORATION 2

Deriving the Quadratic Formula by Completing the Square
Work with a partner.
a. Solve ax2 + bx + c = 0 by completing the square. (Hint: Subtract c from each side, divide each side by a, and then proceed by completing the square.)
Answer:
The given equation is:
ax² + bx + c = 0
Subtract with c on both sides
ax² + bx + c – c = 0 – c
ax² + bx = -c
Divide by a on both sides
x² + \(\frac{b}{a}\)x = –\(\frac{c}{a}\)
Now,
To solve the above equation by using the completing the squares,
Add with \(\frac{b²}{4a²}\) on both sides
So,
x² + \(\frac{b}{a}\) + \(\frac{b²}{4a²}\) = –\(\frac{c}{a}\) + \(\frac{b²}{4a²}\)
(x + \(\frac{b}{2a}\))² = \(\frac{-4c + b²}{4a²}\)
√(x + \(\frac{b}{2a}\))² = √\(\frac{-4c + b²}{4a²}\)
x + \(\frac{b}{2a}\) = \(\frac{\sqrt{-4c + b²}}{2a}\) or x + \(\frac{b}{2a}\) = –\(\frac{\sqrt{-4c + b²}}{2a}\)
x = –\(\frac{b}{2a}\) + \(\frac{\sqrt{b² – 4ac}}{2a}\) or x =  –\(\frac{b}{2a}\) – \(\frac{\sqrt{b² – 4ac}}{2a}\)
Hence, from the above,
We can conclude that the solutions of the given equation are:
–\(\frac{b}{2a}\) + \(\frac{\sqrt{b² – 4ac}}{2a}\) and –\(\frac{b}{2a}\) – \(\frac{\sqrt{b² – 4ac}}{2a}\)

b. Compare this method with the method in Exploration 1. Explain why you think 4a and b2 were chosen in Steps 2 and 3 of Exploration 1.
Answer:
In Exploration 1,
We add 4a to make the expression on the left side of a factor trinomial
We add b² to make the factor trinomial complete i.e., to make the expression on the left side in the form of a² + 2ab + b²

Communicate Your Answer

Question 3.
How can you derive a formula that can be used to write the solutions of any quadratic equation in standard form?
Answer:
The formula can be actually derived using the steps involved in completing the square.
It stems from the fact that any quadratic function or equation of the form
y = ax² + b x + c
can be solved for its roots.

Question 4.
Use the Quadratic Formula to solve each quadratic equation.
a. x2 + 2x – 3 = 0
Answer:
The given expression is:
x² + 2x – 3 = 0
Compare the above equation with
ax² + bx + c = 0
So,
a = 1, b = 2, and c = -3
We know that,
The quadratic formula for fiding the roots is:
x = –\(\frac{b}{2a}\) + \(\frac{\sqrt{b² – 4ac}}{2a}\) and x =  –\(\frac{b}{2a}\) – \(\frac{\sqrt{b² – 4ac}}{2a}\)
So,
x = –\(\frac{b}{2a}\) + \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = –\(\frac{2}{2 (1)}\) + \(\frac{\sqrt{2² – 4 (1) (-3)}}{2 (1)}\)
x = -1 + \(\frac{\sqrt{4 + 12}}{2}\)
x= -1 + 2
x = 1
x =  –\(\frac{b}{2a}\) – \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = –\(\frac{2}{2 (1)}\) – \(\frac{\sqrt{2² – 4 (1) (-3)}}{2 (1)}\)
x = -1 – \(\frac{\sqrt{4 + 12}}{2}\)
x= -1 – 2
x = -3
Hence, from the above,
We can conclude that the solutions of the given equation are: 1 and -3

b. x2 – 4x + 4 = 0
Answer:
The given expression is:
x² – 4x + 4 = 0
Compare the above equation with
ax² + bx + c = 0
So,
a = 1, b = -4, and c = 4
We know that,
The quadratic formula for fiding the roots is:
x = –\(\frac{b}{2a}\) + \(\frac{\sqrt{b² – 4ac}}{2a}\) and x =  –\(\frac{b}{2a}\) – \(\frac{\sqrt{b² – 4ac}}{2a}\)
So,
x = –\(\frac{b}{2a}\) + \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = –\(\frac{-4}{2 (1)}\) + \(\frac{\sqrt{(-4)² – 4 (1) (4)}}{2 (1)}\)
x = 2 + \(\frac{\sqrt{16 – 16}}{2}\)
x= 2
x =  –\(\frac{-4}{2 (1)}\) – \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = 2 – \(\frac{\sqrt{(-4)² – 4 (1) (4)}}{2 (1)1}\)
x = 2 – \(\frac{\sqrt{16 – 16}}{2}\)
x= 2
Hence, from the above,
We can conclude that the solution of the given equation is: 2

c. x2 + 4x + 5 = 0
Answer:
The given expression is:
x² + 4x + 5 = 0
Compare the above equation with
ax² + bx + c = 0
So,
a = 1, b = 4, and c = 5
We know that,
The quadratic formula for fiding the roots is:
x = –\(\frac{b}{2a}\) + \(\frac{\sqrt{b² – 4ac}}{2a}\) and x =  –\(\frac{b}{2a}\) – \(\frac{\sqrt{b² – 4ac}}{2a}\)
So,
x = –\(\frac{b}{2a}\) + \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = –\(\frac{4}{2 (1)}\) + \(\frac{\sqrt{(4)² – 4 (1) (5)}}{2 (1)}\)
x = -2 + \(\frac{\sqrt{16 – 20}}{2}\)
x= -2 + i
x =  –\(\frac{4}{2 (1)}\) – \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = -2 – \(\frac{\sqrt{(4)² – 4 (1) (5)}}{2 (1)}\)
x = -2 – \(\frac{\sqrt{16 – 20}}{2}\)
x= -2 – i
Hence, from the above,
We can conclude that the solutions of the given equation are: -2 ± i

Question 5.
Use the Internet to research imaginary numbers. How are they related to quadratic equations?
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 2
Answer:
An “Imaginary number” is a number that, when squared, has a negative result. Essentially, an imaginary number is the square root of a negative number and does not have a tangible value.  Imaginary numbers are usually denoted by the symbol i
We know that,
√-1 = i

Monitoring Progress

Solve the equation using the Quadratic Formula. Round your solutions to the nearest tenth, if necessary.
Question 1.
x2 – 6x + 5 = 0
Answer:
The given expression is:
x² – 6x + 5 = 0
Compare the above equation with
ax² + bx + c = 0
So,
a = 1, b = -6, and c = 5
We know that,
The quadratic formula for fiding the roots is:
x = –\(\frac{b}{2a}\) + \(\frac{\sqrt{b² – 4ac}}{2a}\) and x =  –\(\frac{b}{2a}\) – \(\frac{\sqrt{b² – 4ac}}{2a}\)
So,
x = –\(\frac{b}{2a}\) + \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = \(\frac{6}{2 (1)}\) + \(\frac{\sqrt{(-6)² – 4 (1) (5)}}{2 (1)}\)
x = 3 + \(\frac{\sqrt{36 – 20}}{2}\)
x= 3 + 2
x = 5
x =  \(\frac{6}{2 (1)}\) – \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = 3 – \(\frac{\sqrt{(-6)² – 4 (1) (5)}}{2 (1)}\)
x = 3 – \(\frac{\sqrt{36 – 20}}{2}\)
x= 3 – 2
x = 1
Hence, from the above,
We can conclude that the solutions of the given equation are: 5 and 1

Question 2.
\(\frac{1}{2}\)x2 + x – 10 = 0
Answer:
The given expression is:
\(\frac{1}{2}\)x² + x – 10 = 0
Multiply with 2 on both sides
So,
x² + 2x – 20 = 0
Compare the above equation with
ax² + bx + c = 0
So,
a = 1, b = 2, and c = -20
We know that,
The quadratic formula for fiding the roots is:
x = –\(\frac{b}{2a}\) + \(\frac{\sqrt{b² – 4ac}}{2a}\) and x =  –\(\frac{b}{2a}\) – \(\frac{\sqrt{b² – 4ac}}{2a}\)
So,
x = –\(\frac{b}{2a}\) + \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = –\(\frac{2}{2 (1)}\) + \(\frac{\sqrt{(2)² + 4 (1) (20)}}{2 (1)}\)
x = -1 + \(\frac{\sqrt{84}}{2}\)
x= 3.5
x =  –\(\frac{2}{2 (1)}\) – \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = -1 – \(\frac{\sqrt{(2)² + 4 (1) (20)}}{2 (1)}\)
x = -1 – \(\frac{\sqrt{84}}{2}\)
x= -4.5
Hence, from the above,
We can conclude that the solutions of the given equation are: 3.5 and -4.5

Question 3.
-3x2 + 2x + 7 = 0
Answer:
The given expression is:
-3x² + 2x + 7 = 0
Compare the above equation with
ax² + bx + c = 0
So,
a = -3, b = 2, and c = 7
We know that,
The quadratic formula for fiding the roots is:
x = –\(\frac{b}{2a}\) + \(\frac{\sqrt{b² – 4ac}}{2a}\) and x =  –\(\frac{b}{2a}\) – \(\frac{\sqrt{b² – 4ac}}{2a}\)
So,
x = –\(\frac{b}{2a}\) + \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = \(\frac{2}{2 (3)}\) + \(\frac{\sqrt{(2)² + 4 (3) (7)}}{2 (-3)}\)
x = \(\frac{1}{3}\) + \(\frac{\sqrt{88}}{-6}\)
x= -1.2
x =  \(\frac{2}{2 (3)}\) – \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = \(\frac{1}{3}\) – \(\frac{\sqrt{(2)² + 4 (3) (7)}}{2 (-3)}\)
x = \(\frac{1}{3}\) + \(\frac{\sqrt{88}}{6}\)
x= 1.8
Hence, from the above,
We can conclude that the solutions of the given equation are: -1.2 and 1.8

Question 4.
4x2 – 4x = -1
Answer:
The given expression is:
4x² – 4x = -1
4x² – 4x + 1 = 0
Compare the above equation with
ax² + bx + c = 0
So,
a = 4, b = -4, and c = 1
We know that,
The quadratic formula for fiding the roots is:
x = –\(\frac{b}{2a}\) + \(\frac{\sqrt{b² – 4ac}}{2a}\) and x =  –\(\frac{b}{2a}\) – \(\frac{\sqrt{b² – 4ac}}{2a}\)
So,
x = –\(\frac{b}{2a}\) + \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = \(\frac{4}{2 (4)}\) + \(\frac{\sqrt{(-4)² – 4 (1) (4)}}{2 (4)}\)
x = \(\frac{1}{2}\) + \(\frac{\sqrt{16 – 16}}{8}\)
x= \(\frac{1}{2}\)
x =  \(\frac{4}{2 (4)}\) – \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = \(\frac{1}{2}\) – \(\frac{\sqrt{(-4)² – 4 (1) (4)}}{2 (4)}\)
x = \(\frac{1}{2}\) – \(\frac{\sqrt{16 – 16}}{8}\)
x= \(\frac{1}{2}\)
Hence, from the above,
We can conclude that the solution of the given equation is: \(\frac{1}{2}\)

Question 5.
WHAT IF?
When were there about 60 wolf breeding pairs?
Answer:
It is given that there are 60 wolf breeding pairs
In Example 2,
The given quadratic equation that used to model the wolf pairs for x years is:
y = 0.20x² + 1.8x – 3
Now,
0.20x² + 1.8x – 3 = 60
0.20x² + 1.8x – 63 =  0
Multiply with 5 on both sides
x² + 9x – 315 = 0
Compare the above equation with
ax² + bx + c = 0
So,
a = 1, b = 9, and c = -315
We know that,
The quadratic formula for fiding the roots is:
x = –\(\frac{b}{2a}\) + \(\frac{\sqrt{b² – 4ac}}{2a}\) and x =  –\(\frac{b}{2a}\) – \(\frac{\sqrt{b² – 4ac}}{2a}\)
So,
x = –\(\frac{b}{2a}\) + \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = –\(\frac{9}{2 (1)}\) + \(\frac{\sqrt{(9)² + 4 (1) (315)}}{2 (1)}\)
x = –\(\frac{9}{2}\) + \(\frac{\sqrt{1,341}}{2}\)
x= 13.80
x = 14
x =  –\(\frac{9}{2 (1)}\) – \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = –\(\frac{9}{2}\) – \(\frac{\sqrt{(9)² + 4 (1) (315)}}{2 (1)}\)
x = –\(\frac{9}{2}\) – \(\frac{\sqrt{1,341}}{2}\)
x= -22.8
= -23
We know that,
Since x represents the number of years, the number of years won’t be negative
So,
x = 14 years
Hence, from the above,
We can conclude that after 14 years, i.e, in 2004, there were 60 wolf breeding pairs

Question 6.
The number y of bald eagle nesting pairs in state x years since 2000 can be modeled by the function y = 0.34x2 + 13.1x + 51.
a. When were there about 160 bald eagle nesting pairs?
Answer:
It is given that
The quadratic equation used to model the number y of bald eagle nesting pairs since 2000 is:
y = 0.34x² + 13.1x + 51
It is given that,
y =160
So,
160 = 0.34x² + 13.1x + 51
0.34x² + 13.1x – 109 = 0
Compare the given equation with
ax² + bx + c = 0
Now,

We know that,
Since x is the number of years, the value of x won’t be negative
So,
x = 7.03 years
x = 7 years (Since the number of years will only be an integer)
Hence, from the above,
We can conclude that after 7 years from 2000, i.e., in 2007, there were 160 bald eagle nesting pairs

b. How many bald eagle nesting pairs were there in 2000?
Answer:
From part (a),
We know that there are 160 bald eagle nesting pairs in 2007
So,
To find the number of bald eagle nesting pairs in 2000,
Subtract the number of bald eagle nesting pairs in 2007 from the difference of 2000 and 2007
So,
The number of bald eagle nesting pairs in 2000 = 160 – (2007 – 2000)
= 160 – 7
= 153
Hence, from the above,
We can conclude that there are 153 bald eagle nesting pairs in 2000

Determine the number of real solutions of the equation.
Question 7.
-x2 + 4x – 4 = 0
Answer:
The given equation is:
-x² + 4x – 4 = 0
Compare the given equation with
ax² + bx + c = 0
So,
a = -1, b = 4, and c = -4
We know that,
If b² – 4ac > 0, then there will be 2 real solutions
If b² – 4ac = 0, then there will be one real solution
If b² – 4ac < 0, then there are no real solutions
So,
b² – 4ac = 4² – 4 (-1) (-4)
= 16 – 16
= 0
Hence, from the above,
We can conclude that there will be only 1 solution for the given equation

Question 8.
6x2 + 2x = -1
Answer:
The given equation is:
6x² + 2x = -1
6x² + 2x + 1 = 0
Compare the given equation with
ax² + bx + c = 0
So,
a = 6, b = 2, and c = 1
We know that,
If b² – 4ac > 0, then there will be 2 real solutions
If b² – 4ac = 0, then there will be one real solution
If b² – 4ac < 0, then there are no real solutions
So,
b² – 4ac = 2² – 4 (6) (1)
= 4 – 24
= -20
Hence, from the above,
We can conclude that there are no real solutions for the given equation

Question 9.
\(\frac{1}{2}\)x2 = 7x – 1
Answer:
The given equation is:
\(\frac{1}{2}\)x² = 7x – 1
\(\frac{1}{2}\)x² – 7x + 1= 0
Multiply with 2 on both sides
So,
x² – 14x + 2 = 0
Compare the given equation with
ax² + bx + c = 0
So,
a = 1, b = -14, and c = 2
We know that,
If b² – 4ac > 0, then there will be 2 real solutions
If b² – 4ac = 0, then there will be one real solution
If b² – 4ac < 0, then there are no real solutions
So,
b² – 4ac = (-14)² – 4 (1) (2)
= 196 – 8
= 188
Hence, from the above,
We can conclude that there are 2 solutions for the given equation

Find the number of x-intercepts of the graph of the function.
Question 10.
y = -x2 + x – 6
Answer:
The given equation is:
y = -x² + x – 6
-x² + x – 6 = 0
Compare the given equation with
ax² + bx + c = 0
So,
a = -1, b = 1, and c = -6
We know that,
If b² – 4ac > 0, then there will be 2 x- intercepts
If b² – 4ac = 0, then there will be one x-intercept
If b² – 4ac < 0, then there are no x-intercepts
So,
b² – 4ac = 1² – 4 (-1) (-6)
= 1 – 24
= -23
Hence, from the above,
We can conclude that there are no x-intercepts for the given equation

Question 11.
y = x2 – x
Answer:
The given equation is:
y = x² – x
x² – x = 0
Compare the given equation with
ax² + bx + c = 0
So,
a = 1, b = -1, and c = 0
We know that,
If b² – 4ac > 0, then there will be 2 x- intercepts
If b² – 4ac = 0, then there will be one x-intercept
If b² – 4ac < 0, then there are no x-intercepts
So,
b² – 4ac = (-1)² – 4 (1) (0)
= 1 – 0
= 1
Hence, from the above,
We can conclude that there are 2 x-intercepts for the given equation

Question 12.
f(x) = x2 + 12x + 36
Answer:
The given equation is:
f (x) = x² + 12x + 36
x² + 12x + 36 = 0
Compare the given equation with
ax² + bx + c = 0
So,
a = 1, b = 12, and c = 36
We know that,
If b² – 4ac > 0, then there will be 2 x- intercepts
If b² – 4ac = 0, then there will be one x-intercept
If b² – 4ac < 0, then there are no x-intercepts
So,
b² – 4ac = 12² – 4 (1) (36)
= 144 – 144
= 0
Hence, from the above,
We can conclude that there is only 1 x-intercept for the given equation

Solve the equation using any method. Explain your choice of method.
Question 13.
x2 + 11x – 12 = 0
Answer:
The given equation is:
x² + 11x – 12 = 0
By observing the above equation, we can easily tell the factors
So,
We can solve the above equation by using the “Factor method”
So,
x² + 11x – 12 = 0
x² + 12x – x – 12 = 0
x (x + 12) – 1 (x + 12) = 0
(x – 1) (x + 2) = 0
x – 1 = 0 or x + 12 = 0
x = 1 or x = -12
Hence, from the above,
We can conclude that the solutions for the given equation are: 1 and -12

Question 14.
9x2 – 5 = 4
Answer:
The given equation is:
9x² – 5 = 4
9x² = 4 + 5
9x² = 9
x² = \(\frac{9}{9}\)
x² = 1
The above equation is in the form of
x² = d
So,
We can solve the above equation by using the “Square root” method
So,
√x² = √1
x = 1 or x = -1
Hence, from the above,
We can conclude that the solutions for the given equation are: 1 and -1

Question 15.
5x2 – x – 1 = 0
Answer:
The given equation is:
5x² – x – 1 = 0
When we observe the above equation,
We can say that we can’t obtain the factors easily
So,
Solve the above equation by using the quadratic formula
Now,
Compare the given equation with
ax² + bx + c = 0
So,
a = 5, b = -1, and c = -1
So,
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{1 ± \sqrt{(-1)² + 4 (1) (5)}}{2 (5)}\)
x = \(\frac{1 ± \sqrt{21}}{10}\)
Hence, from the above,
We can conclude that the solutions of the given equation are: \(\frac{1 ± \sqrt{21}}{10}\)

Question 16.
x2 = 2x – 5
Answer:
The given equation is:
x² = 2x – 5
x² – 2x = 5
When we observe the equation,
We can say that the expression on the left side can e solved easily by using the completing the squares method
So,
We can use the “Completing the squares” method to solve the above equation
So,
Add 1 on both sides
x² – 2x + 1 = 5 + 1
(x – 1)² = 6
√(x – 1)² = √6
x – 1 = √6 or x – 1 = -√6
x = √6 + 1 or x = -√6 + 1
Hence, from the above,
We can conclude that the solutions for the given equation are: √6 + 1 and -√6 + 1

Solving Quadratic Equations Using the Quadratic Formula 9.5 Exercises

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
What formula can you use to solve any quadratic equation? Write the formula.
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 1

Question 2.
VOCABULARY
In the Quadratic Formula, what is the discriminant? What does the value of the discriminant determine?
Answer:
The quadratic formula is:
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
In the above quadratic formula,
The discriminant is: b² – 4ac
We know that,
If b² – 4ac > 0, then there will be 2 real solutions
If b² – 4ac = 0, then there will be one real solution
If b² – 4ac < 0, then there are no real solutions

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, write the equation in standard form. Then identify the values of a, b, and c that you would use to solve the equation using the Quadratic Formula.
Question 3.
x2 = 7x
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 3

Question 4.
x2 – 4x = -12
Answer:
The given equation is:
x² – 4x = -12
Method 1:
Add with 12 on both sides
x² – 4x + 12 = 12 – 12
x² – 4x + 12 = 0
Compare the above equation with
ax² + bx + c = 0
So,
a = 1, b = -4, and c = 12
Method 2:
Add with 4x on both sides
x² – 4x + 4x = -12 + 4x
x² = -12 + 4x
Subtract with x² on both sides
x² – x² = -12 + 4x – x²
-12 + 4x – x² = 0
Compare the given equation with
ax² + bx + c = 0
So,
a = -1, b = 4, and c = -12

Question 5.
-2x2 + 1 = 5x
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 5

Question 6.
3x + 2 = 4x2
Answer:
The given equation is:
3x + 2 = 4x²
Method 1:
Subtract with 2 on both sides
3x + 2 – 2 = 4x² – 2
Subtract with 3x on both sides
3x – 3x = 4x² – 2 – 3x
4x² – 3x – 2 = 0
Compare the above equation with
ax² + bx + c = 0
So,
a = 4, b = -3, and c = -2
Method 2:
Subtract with 4x² on both sides
3x + 2 – 4x² = 4x² – 4x²
-4x² + 3x + 2 = 0
Compare the given equation with
ax² + bx + c = 0
So,
a = -4, b = 3, and c = 2

Question 7.
4 – 3x = -x2 + 3x
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 7

Question 8.
-8x – 1 = 3x2 + 2
Answer:
The given equation is:
-8x – 1 = 3x² + 2
-8x -1 – 2 = 3x²
-8x – 3 = 3x²
Method 1:
Add with 3 on both sides
-8x – 3 + 3 = 3x² + 3
-8x = 3x² + 3
Add with 8x on both sides
8x – 8x = 3x² + 3 + 8x
3x² + 8x + 3 = 0
Compare the above equation with
ax² + bx + c = 0
So,
a = 3, b = 8, and c = 3
Method 2:
Subtract with 3x² on both sides
-8x – 3 – 3x² = 3x² – 3x²
-3x² – 8x – 3 = 0
3x² + 8x + 3 = 0
Compare the given equation with
ax² + bx + c = 0
So,
a = 3, b = 8, and c = 3

In Exercises 9–22, solve the equation using the Quadratic Formula. Round your solutions to the nearest tenth, if necessary.
Question 9.
x2 – 12x + 36 = 0
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 9

Question 10.
x2 + 7x + 16 = 0
Answer:
The given equation is:
x² + 7x + 16 = 0
Compare the given equation with
ax² + bx + c = 0
So,
a = 1, b = 7, and c = 16
So,
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{-7 ± \sqrt{(7)² – 4 (1) (16)}}{2 (1)}\)
x = \(\frac{-7 ± \sqrt{-15}}{2}\)
x = \(\frac{-7 ± 15i}{2}\)
Hence, from the above,
We can conclude that the solutions of the given equation are: \(\frac{-7 ± 15i}{2}\)

Question 11.
x2 – 10x – 11 = 0
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 11

Question 12.
2x2 – x – 1 = 0
Answer:
The given equation is:
2x² – x – 1 = 0
Compare the given equation with
ax² + bx + c = 0
So,
a = 2, b = -1, and c = -1
So,
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{1 ± \sqrt{(-1)² + 4 (1) (2)}}{2 (2)}\)
x = \(\frac{1 ± \sqrt{9}}{4}\)
x = \(\frac{1 ± 3}{4}\)
x = \(\frac{1 + 3}{4}\) or x = \(\frac{1 – 3}{4}\)
x = 1 or x = –\(\frac{1}{2}\)
Hence, from the above,
We can conclude that the solutions of the given equation are: 1 and –\(\frac{1}{2}\)

Question 13.
2x2 – 6x + 5 = 0
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 13

Question 14.
9x2 – 6x + 1 = 0
Answer:
The given equation is:
9x² – 6x + 1 = 0
Compare the given equation with
ax² + bx + c = 0
So,
a = 9, b = -6, and c = 1
So,
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{6 ± \sqrt{(-6)² – 4 (1) (9)}}{2 (9)}\)
x = \(\frac{6 ± \sqrt{0}}{18}\)
x = \(\frac{6 + 0}{18}\) or x = \(\frac{6 – 0}{18}\)
x = \(\frac{6}{18}\) or x = \(\frac{6}{18}\)
x = \(\frac{1}{3}\) or x = \(\frac{1}{3}\)
Hence, from the above,
We can conclude that the solution of the given equation is: \(\frac{1}{3}\)

Question 15.
6x2 – 13x = -6
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 15

Question 16.
-3x2 + 6x = 4
Answer:
The given equation is:
-3x² + 6x = 4
-3x² + 6x – 4 = 0
3x² – 6x + 4 = 0
Compare the given equation with
ax² + bx + c = 0
So,
a = 3, b = -6, and c = 4
So,
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{6 ± \sqrt{(-6)² – 4 (3) (4)}}{2 (3)}\)
x = \(\frac{6 ± \sqrt{-12}}{6}\)
x = \(\frac{6 ± 12i}{6}\)
x = 1 ± 2i
Hence, from the above,
We can conclude that the solutions of the given equation are: 1 ±2i

Question 17.
1 – 8x = -16x2
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 17

Question 18.
x2 – 5x + 3 = 0
Answer:
The given equation is:
x² – 5x + 3 = 0
Compare the given equation with
ax² + bx + c = 0
So,
a = 1, b = -5, and c = 3
So,
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{5 ± \sqrt{(-5)² – 4 (1) (3)}}{2 (1)}\)
x = \(\frac{5 ± \sqrt{13}}{2}\)
x = \(\frac{5 ± 3.6}{2}\)
x = \(\frac{5 + 3.6}{2}\) or x = \(\frac{5 – 3.6}{2}\)
x = 4.3 or x = 0.7
Hence, from the above,
We can conclude that the solutions of the given equation are: 4.3 and 0.7

Question 19.
x2 + 2x = 9
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 19

Question 20.
5x2 – 2 = 4x
Answer:
The given equation is:
5x² – 2 = 4x
5x² – 4x – 2 = 0
Compare the given equation with
ax² + bx + c = 0
So,
a = 5, b = -4, and c = -2
So,
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{4 ± \sqrt{(-4)² + 4 (5) (2)}}{2 (5}\)
x = \(\frac{4 ± \sqrt{56}}{10}\)
x = \(\frac{4 ± 7.4}{10}\)
x = \(\frac{4 + 7.4}{10}\) or x = \(\frac{4 – 7.4}{10}\)
x = 1.14 or x = -0.34
Hence, from the above,
We can conclude that the solutions of the given equation are:1.14 and -0.34

Question 21.
2x2 + 9x + 7 = 3
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 21

Question 22.
8x2 + 8 = 6 – 9x
Answer:
The given equation is:
8x² + 8 = 6 – 9x
8x² + 9x + 8 – 6 = 0
8x² + 9x + 2 = 0
Compare the given equation with
ax² + bx + c = 0
So,
a = 8, b = 9, and c = 2
So,
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{-9 ± \sqrt{(9)² – 4 (8) (2)}}{2 (8)}\)
x = \(\frac{-9 ± \sqrt{17}}{16}\)
x = \(\frac{-9 ± 4.1}{16}\)
x = \(\frac{-9 + 4.1}{16}\) or x = \(\frac{-9 – 4.1}{16}\)
x = -0.3 or x = -0.8
Hence, from the above,
We can conclude that the solutions of the given equation are: -0.3 and -0.8

Question 23.
MODELING WITH MATHEMATICS
A dolphin jumps out of the water, as shown in the diagram. The function h = -16t2 + 26t models the height h (in feet) of the dolphin after t seconds. After how many seconds is the dolphin at a height of 5 feet?
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 3
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 23

Question 24.
MODELING WITH MATHEMATICS
The amount of trout y (in tons) caught in a lake from 1995 to 2014 can be modeled by the equation y = -0.08x2 + 1.6x + 10, where x is the number of years since 1995.
a. When were about 15 tons of trout caught in the lake?
Answer:
It is given that the quadratic equation used to model the amount of trout from 1995 to 2014 is:
y = -0.08x² + 1.6x + 10
It is given that
y = 15
So,
15 = -0.08x² + 1.6x + 10
-0.08x² + 1.6x – 5 = 0
Compare the given equation with
ax² + bx + c = 0
Now,

From the above,
We can observe that
The values of x are: 4 and 16
Now,
For 15 tons of trout caught in the lake, the number of years will be less
Hence, from the above,
We can conclude that after 4 years, i.e., in 1999, 15 ton of trout were caught in the lake

b. Do you think this model can be used to determine the amounts of trout caught in future years? Explain your reasoning.
Answer:
Yes, we can use this model to determine the amounts of trout caught in future years but the maximum of trout can be caught only in 16 years

In Exercises 25–30, determine the number of real solutions of the equation.
Question 25.
x2 – 6x + 10 = 0
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 25

Question 26.
x2 – 5x – 3 = 0
Answer:
The given equation is:
x² – 5x – 3 = 0
Compare the given equation with
ax² + bx + c = 0
So,
a = 1, b = -5, and c = -3
We know that,
If b² – 4ac > 0, then there will be 2 real solutions
If b² – 4ac = 0, then there will be one real solution
If b² – 4ac < 0, then there are no real solutions
So,
b² – 4ac = (-5)² – 4 (1) (-3)
= 25 + 12
= 37
Hence, from the above,
We can conclude that there will be 2 solutions for the given equation

Question 27.
2x2 – 12x = -18
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 27

Question 28.
4x2 = 4x – 1
Answer:
The given equation is:
4x² = 4x – 1
4x² – 4x + 1 = 0
Compare the given equation with
ax² + bx + c = 0
So,
a = 4, b = -4, and c = 1
We know that,
If b² – 4ac > 0, then there will be 2 real solutions
If b² – 4ac = 0, then there will be one real solution
If b² – 4ac < 0, then there are no real solutions
So,
b² – 4ac = (-4)² – 4 (1) (4)
= 16 – 16
= 0
Hence, from the above,
We can conclude that there will be only 1 solution for the given equation

Question 29.
–\(\frac{1}{4}\)x2 + 4x = -2
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 29

Question 30.
-5x2 + 8x = 9
Answer:
The given equation is:
-5x² + 8x = 9
-5x² + 8x – 9 = 0
Compare the given equation with
ax² + bx + c = 0
So,
a = -5, b = 8, and c = -9
We know that,
If b² – 4ac > 0, then there will be 2 real solutions
If b² – 4ac = 0, then there will be one real solution
If b² – 4ac < 0, then there are no real solutions
So,
b² – 4ac = 8² – 4 (-5) (-9)
= 64 – 180
= -116
Hence, from the above,
We can conclude that there will be no  real solutions for the given equation

In Exercises 31–36, find the number of x-intercepts of the graph of the function.
Question 31.
y = x2 + 5x – 1
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 31

Question 32.
y = 4x2 + 4x + 1
Answer:
The given equation is:
y = 4x² + 4x + 1
4x² + 4x + 1 = 0
Compare the given equation with
ax² + bx + c = 0
So,
a = 4, b = 4, and c = 1
We know that,
If b² – 4ac > 0, then there will be 2 x- intercepts
If b² – 4ac = 0, then there will be one x-intercept
If b² – 4ac < 0, then there are no x-intercepts
So,
b² – 4ac = 4² – 4 (1) (4)
= 16 – 16
= 0
Hence, from the above,
We can conclude that there will be 1 x-intercept for the given equation

Question 33.
y = -6x2 + 3x – 4
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 33

Question 34.
y = -x2 + 5x + 13
Answer:
The given equation is:
y = -x² + 5x + 13
-x² + 5x + 13 = 0
Compare the given equation with
ax² + bx + c = 0
So,
a = -1, b = 5, and c = 13
We know that,
If b² – 4ac > 0, then there will be 2 x- intercepts
If b² – 4ac = 0, then there will be one x-intercept
If b² – 4ac < 0, then there are no x-intercepts
So,
b² – 4ac = 5² – 4 (-1) (13)
= 25 + 52
= 77
Hence, from the above,
We can conclude that there are 2 x-intercepts for the given equation

Question 35.
f(x) = 4x2 + 3x – 6
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 35

Question 36.
f(x) = 2x2 + 8x + 8
Answer:
The given equation is:
y = 2x² + 8x + 8
2x² +8x + 8 = 0
Compare the given equation with
ax² + bx + c = 0
So,
a = 2, b = 8, and c = 8
We know that,
If b² – 4ac > 0, then there will be 2 x- intercepts
If b² – 4ac = 0, then there will be one x-intercept
If b² – 4ac < 0, then there are no x-intercepts
So,
b² – 4ac = 8² – 4 (2) (8)
= 64 – 64
= 0
Hence, from the above,
We can conclude that there is only 1 x-intercept for the given equation

In Exercises 37–44, solve the equation using any method. Explain your choice of method.
Question 37.
-10x2 + 13x = 4
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 37

Question 38.
x2 – 3x – 40 = 0
Answer:
The given equation is:
x² – 3x – 40 = 0
When we observe the above equation,
We can easily say the factors
So,
We can use the “Factor method” to solve the above equation
So,
x² – 8x + 5x – 40 = 0
x (x + 5) – 8 (x + 5) = 0
(x – 8) (x + 5) = 0
x – 8 = 0 or x + 5 = 0
x = 8 or x = -5
Hence, from the above,
We can conclude that the solutions of the given equation are: 8 and -5

Question 39.
x2 + 6x = 5
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 39

Question 40.
-5x2 = -25
Answer:
The given equation is:
-5x² = -25
5x² = 25
x² = \(\frac{25}{5}\)
x² = 5
When we observe the above equation,
We can say that the equation is in the form of
x² = d
So,
We can use the “Square root” method to solve this equation
So,
√x² = √5
x = ±√5
Hence, from the above,
We can conclude that the solution for the above equation are: ±√5

Question 41.
x2 + x – 12 = 0
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 41

Question 42.
x2 – 4x + 1 = 0
Answer:
The given equation is:
x² – 4x + 1 = 0
When we observe the above equation,
We can say that the expression on the left side can be solved by “Completing the squares”
So,
We can use the “Completing the squares” method to solve the above equation
So,
Add with 4 on both sides
x² – 4x + 4 = -1 + 4
(x – 2)² = 3
√(x – 2)² = √3
x – 2 = −√3 or x – 2 = √3
x = -√3 + 2 or x = √3 + 2
Hence, from the above,
We can conclude that the solutions for the given equation are: √3 + 2 and -√3 + 2

Question 43.
4x2 – x = 17
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 43

Question 44.
x2 + 6x + 9 = 16
Answer:
The given equation is:
x² + 6x + 9 = 16
x² + 6x = 16 – 9
x² + 6x = 7
When we observe the above equation,
We can say that the expression on the left side can be solved by “Completing the squares”
So,
We can use the “Completing the squares” method to solve the above equation
So,
Add with 9 on both sides
x² + 6x + 9 = 7 + 9
(x + 3)² = 16
√(x + 3)² = √16
x + 3 = −4 or x + 3 = 4
x = -4 – 3 or x = 4 – 3
x = -7 or x = 1
Hence, from the above,
We can conclude that the solutions for the given equation are: 1 and -7

Question 45.
ERROR ANALYSIS
Describe and correct the error in solving the equation 3x2 – 7x – 6 = 0 using the Quadratic Formula
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 4
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 45

Question 46.
ERROR ANALYSIS
Describe and correct the error in solving the equation -2x2 + 9x = 4 using the Quadratic Formula.
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 5
Answer:
The given equation is:
-2x² + 9x = 4
-2x² + 9x – 4 = 0
2x² – 9x + 4 = 0
Compare the given equation with
ax² + bx + c = 0
So,
a = 2, b = -9, and c = 4
So,
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{9 ± \sqrt{(-9)² – 4 (2) (4)}}{2 (2}\)
x = \(\frac{9 ± \sqrt{49}}{4}\)
x = \(\frac{9 ± 7}{4}\)
x = \(\frac{9 + 7}{4}\) or x = \(\frac{9 – 7}{4}\)
x = 4 or x = \(\frac{1}{2}\)
Hence, from the above,
We can conclude that the solutions of the given equation are: 4 and \(\frac{1}{2}\)

Question 47.
MODELING WITH MATHEMATICS
A fountain shoots a water arc that can be modeled by the graph of the equation y = -0.006x2 + 1.2x + 10, where x is the horizontal distance (in feet) from the river’s north shore and y is the height (in feet) above the river. Does the water arc reach a height of 50 feet? If so, how far from the north shore is the water arc 50 feet above the water?
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 6
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 47

Question 48.
MODELING WITH MATHEMATICS
Between the months of April and September, the number y of hours of daylight per day in Seattle, Washington, can be modeled by y = -0.00046x2 + 0.076x + 13, where x is the number of days since April 1.
a. Do any of the days between April and September in Seattle have 17 hours of daylight? If so, how many?
Answer:

b. Do any of the days between April and September in Seattle have 14 hours of daylight? If so, how many?
Answer:

Question 49.
MAKING AN ARGUMENT
Your friend uses the discriminant of the equation 2x2 – 5x – 2 = -11 and determines that the equation has two real solutions. Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 49

Question 50.
MODELING WITH MATHEMATICS
The frame of the tent shown is defined by a rectangular base and two parabolic arches that connect the opposite corners of the base. The graph of y = -0.18x2 + 1.6x models the height y (in feet) of one of the arches x feet along the diagonal of the base. Can a child who is 4 feet tall walk under one of the arches without having to bend over? Explain.
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 7
Answer:


Hence, from the above,
We can conclude that can not walk without having to bend over

MATHEMATICAL CONNECTIONS In Exercises 51 and 52, use the given area A of the rectangle to find the value of x. Then give the dimensions of the rectangle.
Question 51.
A = 91 m2
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 8
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 51.1
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 51.2

Question 52.
A = 209 ft2
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 9
Answer:
The given figure is:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 9
We know that,
Area = Length × Width
So,
From the given figure,
209 = (4x – 5) (4x + 3)
16x² + 12x – 20x – 15 = 209
16x² – 8x – 15 – 209 = 0
16x² – 8x – 224 = 0
Divide by 8 on both sides
So,
2x² – x – 28 = 0
2x² – 8x + 7x – 28 = 0
2x ( x – 4) + 7 (x – 4) = 0
(2x + 7) (x – 4) = 0
2x + 7 = 0 or x – 4 = 0
2x = -7 or x = 4
x = –\(\frac{7}{2}\) or x = 4
We know that,
The dimensions of a rectangle won’t be negative
So,
The value of x is: 4
Now,
The length of rectangle = 4x + 3
= 4 (4) + 3
= 16 + 3 = 19 ft
The width of the rectangle = 4x – 5
= 4 (4) – 5
= 16 – 5
= 11 ft
Hence, from the above,
We can conclude that
The length of the rectangle is: 19 ft
The width of the rectangle is: 11 ft

COMPARING METHODS In Exercises 53 and 54, solve the equation by (a) graphing, (b) factoring, and (c) using the Quadratic Formula. Which method do you prefer? Explain your reasoning.
Question 53.
x2 + 4x + 4 = 0
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 53.1
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 53.2

Question 54.
3x2 + 11x + 6 = 0
Answer:
The given equation is:
3x² + 11x + 6 = 0
a) By using graphing method:

Hence, from the above,
We can conclude that the solutions for the given equation are: -3 and -0.6
b) By using factoring method:
The given equation is:
3x² + 11x + 6 = 0
3x² + 9x + 2x + 6 = 0
3x (x + 3) + 2 (x + 3) = 0
(3x + 2) (x + 3) = 0
3x + 2 = 0 or x + 3 = 0
3x = -2 or x = -3
x = –\(\frac{2}{3}\) or x = -3
Hence, from the above,
We can conclude that the solutions for the given equation are: -3 and –\(\frac{2}{3}\)
c) By using the quadratic formula:
The given equation is:
3x² + 11x + 6 = 0
Compare the given equation with
ax² + bx + c = 0
So,
a = 3, b = 11, and c = 6
So,
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{-11 ± \sqrt{(11)² – 4 (3) (6)}}{2 (3}\)
x = \(\frac{-11 ± \sqrt{49}}{6}\)
x = \(\frac{-11 ± 7}{6}\)
x = \(\frac{-11 + 7}{6}\) or x = \(\frac{-11 – 7}{6}\)
x = –\(\frac{2}{3}\) or x = -3
Hence, from the above,
We can conclude that the solutions of the given equation are: -3 and –\(\frac{2}{3}\)
Hence, from the above 3 methods,
We can conclude that we can find the solutions for the given equation easily by using the factoring method

Question 55.
REASONING
How many solutions does the equation ax2 + bx + c = 0 have when a and c have different signs? Explain your reasoning.
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 55

Question 56.
REASONING
When the discriminant is a perfect square, are the solutions of ax2 + bx + c = 0 rational or irrational? (Assume a, b, and c are integers.) Explain your reasoning.
Answer:
It is given that
The discriminant is a perfect square
We know that,
Discriminant = b² – 4ac
So,
A perfect square means a number that is greater than zero
So,
b² – 4ac > 0
Hence, from the above,
We can conclude that the solutions of ax² + bx + c = 0 are rational

REASONING In Exercises 57–59, give a value of c for which the equation has (a) two solutions, (b) one solution, and (c) no solutions.
Question 57.
x2 – 2x + c = 0
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 57.1
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 57.2

Question 58.
x2 – 8x + c = 0
Answer:
The given equatin is:
x² – 8x + c = 0
Compare the given equation with
ax² + bx + c = 0
So,
a = 1 and b = -8
We know that,
Discriminant (d) = b² – 4ac
If d > 0, then the equation have 2 real solutions
If d = 0, then the equation has one solution
If d < 0, then the equation have no real solutions
a) For two solutions:
b² – 4ac > 0
(-8)² – 4 (1) c > 0
64 – 4c > 0
64 > 4c
\(\frac{64}{4}\) > c
16 > c
c < 16
b) For one solution:
b² – 4ac = 0
(-8)² – 4 (1) c = 0
64 – 4c = 0
64 = 4c
\(\frac{64}{4}\) = c
16 = c
c = 16
c) For no solutions:
b² – 4ac < 0
(-8)² – 4 (1) c < 0
64 – 4c < 0
64 < 4c
\(\frac{64}{4}\) < c
16 < c
c > 16

Question 59.
4x2 + 12x + c = 0
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 59.1
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 59.2

Question 60.
REPEATED REASONING
You use the Quadratic Formula to solve an equation.
a. You obtain solutions that are integers. Could you have used factoring to solve the equation? Explain your reasoning.
Answer:
Yes, the solutions correspond to zeros. If solutions are p and q, we can write the equation in the form
(x – p) (x – q) = 0
So,
The equation is factorized with the solutions p and q
Example:
If  the solutions are
p = 2 and q = -3, then
(x – 2) (x + 3) = 0

b. You obtain solutions that are fractions. Could you have used factoring to solve the equation? Explain your reasoning.
Answer:
Yes, the reason is the same as in part (a)
Example:
If the solutions are:
p = \(\frac{1}{2}[/altex] and q = –[latex]\frac{4}{3}\)
then,
(2x – 1) (3x + 4) = 0

c. Make a generalization about quadratic equations with rational solutions.
Answer:
The quadratic equations with the rational solutions can be easily solved using the “factoring” method of the quadratic equation

Question 61.
MODELING WITH MATHEMATICS
The fuel economy y(in miles per gallon) of a car can be modeled by the equation y = -0.013x2 + 1.25x + 5.6, where 5 ≤ x ≤ 75 and x is the speed (in miles per hour) of the car. Find the speed(s) at which you can travel and have a fuel economy of 32 miles per gallon.
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 61

Question 62.
MODELING WITH MATHEMATICS
The depth d (in feet) of a river can be modeled by the equation d = -0.25t2 + 1.7t + 3.5, where 0 ≤ t ≤ 7 and t is the time (in hours) after a heavy rain begins. When is the river 6 feet deep?
Answer:

ANALYZING EQUATIONS In Exercises 63–68, tell whether the vertex of the graph of the function lies above, below, or on the x-axis. Explain your reasoning without using a graph.
Question 63.
y = x2 – 3x + 2
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 63

Question 64.
y = 3x2 – 6x + 3
Answer:
The given equation is:
y = 3x² – 6x + 3
So,
3x² – 6x + 3 = 0
Compare the given equation with
ax² + bx + c = 0
So,
a = 3, b = -6, and c = 3
Now,
b² – 4ac = (-6)² – 4 (3) (3)
= 36 – 36
= 0
From the above,
We can observe that the discriminant is equal to 0
Hence, from the above,
We can conclude that the vertex of the given function lies on x-axis

Question 65.
y = 6x2 – 2x + 4
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 65

Question 66.
y = -15x2 + 10x – 25
Answer:
The given equation is:
y = -15x² + 10x – 25
So,
-15x² + 10x – 25 = 0
15x² – 10x + 25 = 0
Compare the given equation with
ax² + bx + c = 0
So,
a = 15, b = -10, and c = 25
Now,
b² – 4ac = (-10)² – 4 (15) (25)
= 100 – 1,500
= -1,400
From the above,
We can observe that the discriminant is less than 0
Hence, from the above,
We can conclude that the graph of y has no x-intercepts and also because a > 0, the graph opens up and has a minimum value.
So,
The vertex of the graph lies above the x-axis

Question 67.
f(x) = -3x2 – 4x + 8
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 67

Question 68.
f(x) = 9x2 – 24x + 16
Answer:
The given equation is:
f (x) = 9x² – 24x + 16
So,
9x² – 24x + 16 = 0
Compare the given equation with
ax² + bx + c = 0
So,
a = 9, b = -24, and c = 16
Now,
b² – 4ac = (-24)² – 4 (9) (16)
= 576 – 576
= 0
From the above,
We can observe that the discriminant is equal to 0
Hence, from the above,
We can conclude that the vertex of the given function lies on the x-axis

Question 69.
REASONING
NASA creates a weightless environment by flying a plane in a series of parabolic paths. The height h (in feet) of a plane after t seconds in a parabolic flight path can be modeled by h = -11t2 + 700t + 21,000. The passengers experience a weightless environment when the height of the plane is greater than or equal to 30,800 feet. For approximately how many seconds do passengers experience weightlessness on such a flight? Explain.
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 10
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 69

Question 70.
WRITING EQUATIONS
Use the numbers to create a quadratic equation with the solutions x = -1 and x = –\(\frac{1}{4}\)
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 11
Answer:
The solutions to the given equation are:
x = -1 and x = –\(\frac{1}{4}\)
We know that,
The representation of the equation with the solutions are:
(x – p) (x – q) = 0
So,
(x + 1) (x + \(\frac{1}{4}\)) = 0
(x + 1) (4x + 1) = 0
4x² + x + 4x + 1 = 0
4x² + 5x + 1 = 0
Hence, from the above,
We can conclude that the missing numbers of the given equation are: 4, 5, and 1

Question 71.
PROBLEM SOLVING
A rancher constructs two rectangular horse pastures that share a side, as shown. The pastures are enclosed by 1050 feet of fencing. Each pasture has an area of 15,000 square feet.
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 12
a. Show that y = 350 – \(\frac{4}{3}\)x.
b. Find the possible lengths and widths of each pasture.
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 71.1
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 71.2

Question 72.
PROBLEM SOLVING
A kicker punts a football from a height of 2.5 feet above the ground with an initial vertical velocity of 45 feet per second.
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 13
a. Write an equation that models this situation using the function h = -16t2 + v0t + s0, where h is the height (in feet) of the football, t is the time (in seconds) after the football is punted, v0 is the initial vertical velocity (in feet per second), and s0 is the initial height (in feet).
b. The football is caught 5.5 feet above the ground, as shown in the diagram. Find the amount of time that the football is in the air.
Answer:

Question 73.
CRITICAL THINKING
The solutions of the quadratic equation ax2 + bx + c = 0 are x = \(\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}\) and x = \(\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}\). Find the mean of the solutions. How is the mean of the solutions related to the graph of y = ax2 + bx + c? Explain.
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 73

Question 74.
HOW DO YOU SEE IT?
Match each graph with its discriminant. Explain your reasoning.
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 14
a. b2 – 4ac > 0
Answer:
We know that,
If b² – 4ac > 0, then the quadratic equation will have 2 real roots
So,
When we observe the graphs, graph C matches the b² – 4ac > 0
SInce a < 0, the quadratic equation will have the parabola closed down and have the maximum value
Hence,
The graph of b² – 4ac > 0 will have the maximum value

b. b2 – 4ac = 0
Answer:
We know that,
If b² – 4ac = 0, then the quadratic equation will have 1 real root
So,
When we observe the graphs, graph A matches the b² – 4ac = 0
SInce a < 0, the quadratic equation will have the parabola closed down and have the maximum value
Hence,
The graoh of b² – 4ac = 0 will have the maximum value

c. b2 – 4ac < 0
Answer:
We know that,
If b² – 4ac< 0, then the quadratic equation will have no real roots
So,
When we observe the graphs, graph B matches the b² – 4ac < 0
SInce a > 0, the quadratic equation will have the parabola opens up and have the minimum imaginary value
Hence,
The graph of b² – 4ac < 0 will have the minimum imaginary value

Question 75.
CRITICAL THINKING
You are trying to hang a tire swing. To get the rope over a tree branch that is 15 feet high, you tie the rope to a weight and throw it over the branch. You release the weight at a height s0 of 5.5 feet. What is the minimum initial vertical velocity v0 needed to reach the branch? (Hint: Use the equation h = -16t2 + v0 t + s0)
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 75

Question 76.
THOUGHT-PROVOKING
Consider the graph of the standard form of a quadratic function y = ax2 + bx + c. Then consider the Quadratic Formula as given by
x = \(-\frac{b}{2 a} \pm \frac{\sqrt{b^{2}-4 a c}}{2 a}\)
Write a graphical interpretation of the two parts of this formula.
Answer:

Question 77.
ANALYZING RELATIONSHIPS
Find the sum and product of \(\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}\) and \(\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}\). Then write a quadratic equation whose solutions have a sum of 2 and a product of \(\frac{1}{2}\).
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 77.1
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 77.2

Question 78.
WRITING A FORMULA
Derive a formula that can be used to find solutions of equations that have the form ax2 + x + c = 0. Use your formula to solve -2x2 + x + 8 = 0.
Answer:
The given equation is:
ax² + x + c = 0
Compare the given equation with
ax² + bx + c = 0
So,
a = a, b = 1, and c = c
So,
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{-1 ± \sqrt{1² – 4 (a) (c)}}{2a}\)
x = \(\frac{-1 ± \sqrt{1 – 4ac}}{2a}\)
So,
The solutions of the equation ax² + x + c = 0 are: \(\frac{-1 ± \sqrt{1 – 4ac}}{2a}\)
Now,
The example of ax² + x + c is:
-2x² + x + 8 = 0
Compare the above equation with
ax² + x + c = 0
So,
a = -2, b = 1, and c = 8
So,
The solutions of the above equation are:
x = \(\frac{-1 ± \sqrt{1 – 4 (-2) (8)}}{2 (-2)}\)
x = \(\frac{-1 ± \sqrt{1 + 64}}{-4}\)
x = \(\frac{-1 ± \sqrt{65}}{-4}\)
x = \(\frac{-1 + \sqrt{65}}{-4}\) or x = \(\frac{-1 – \sqrt{65}}{-4}\)
x = -1.76 or x = 2.26
Hence, from the above,
We can conclude that the solutions of the equation -2x² + x + 8 = 0 are: -1.76 and 2.26

Question 79.
MULTIPLE REPRESENTATIONS
If p is a solution of a quadratic equation ax2 + bx + c = 0, then (x – p) is a factor of ax² + bx + c.
a. Copy and complete the table for each pair of solutions.
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 15
b. Graph the related function for each equation. Identify the zeros of the function.
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 79.1
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 79.2
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 79.3

CRITICAL THINKING In Exercises 80–82, find all values of k for which the equation has (a) two solutions, (b) one solution, and (c) no solutions.
Question 80.
2x2 + x + 3k = 0
Answer:
The given equation is:
2x² + x + 3k = 0
Compare the givene quation with
ax² + bx + c = 0
So,
a = 2, b = 1, and c = 3k
We know that,
If b² – 4ac > 0, then the quadratic equation has 2 soltions
If b² – 4ac = 0, then the quadratic equation has 1 solution
If b² – 4ac < 0, then the quadratic equation has no real solutions
Now,
a) For 2 solutions:
b² – 4ac > 0
1² – 4 (2) (3k) > 0
1 – 24k > 0
1 > 24k
24k < 1
k < \(\frac{1}{24}\)
b) For 1 solution:
b² – 4ac = 0
1² – 4 (2) (3k) = 0
1 – 24k = 0
1 = 24k
24k = 1
k = \(\frac{1}{24}\)
c) For no solutions:
b² – 4ac < 0
1² – 4 (2) (3k) < 0
1 – 24k < 0
1 < 24k
24k > 1
k > \(\frac{1}{24}\)

Question 81.
x2 = 4kx + 36 = 0
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 81.1
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 81.2

Question 82.
kx2 + 5x – 16 = 0
Answer:
The given equation is:
kx² + 5x – 16 = 0
Compare the givene quation with
ax² + bx + c = 0
So,
a = k, b = 5, and c = -16
We know that,
If b² – 4ac > 0, then the quadratic equation has 2 soltions
If b² – 4ac = 0, then the quadratic equation has 1 solution
If b² – 4ac < 0, then the quadratic equation has no real solutions
Now,
a) For 2 solutions:
b² – 4ac > 0
5² – 4 (k) (-16) > 0
25 + 64k > 0
25 > -64k
-64k < 25
k < –\(\frac{25}{64}\)
b) For 1 solution:
b² – 4ac = 0
5² – 4 (k) (-16) = 0
25 + 64k = 0
25 = -64k
-64k = 25
k = –\(\frac{25}{64}\)
c) For no solutions:
b² – 4ac < 0
5² – 4 (k) (-16) < 0
25 + 64k = 0
25 < -64k
-64k > 25
k > –\(\frac{25}{64}\)

Maintaining Mathematical Proficiency

Solve the system of linear equations using any method. Explain why you chose the method.
Question 83.
y = -x + 4
y = 2x – 8
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 83

Question 84.
x = 16 – 4y
3x + 4y = 8
Answer:
The given system of linear equations are:
x = 16 – 4y
3x + 4y = 8
By observeing the given system of linear equations,
We can say that we can solve the linear equations by using the substitution method
So,
x + 4y = 16 —-(1)
3x + 4y = 8 —-(2)

Hence, from the above,
We can conclude that the solution of the gven system of linear equations is: (-4, 5)

Question 85.
2x – y = 7
2x + 7y = 31
Answer:
Big Ideas Math Answers Algebra 1 Chapter 9 Solving Quadratic Equations 9.5 a 85

Question 86.
3x – 2y = -20
x + 1.2y = 6.4
Answer:
The given system of linear equations are:
3x – 2y = -20 —-(1)
x + 1.2y = 6.4 —(2)
By observing the above system of linear equations,
We can say that the given system of linear equations can be solved by the elimination method
So,

Hence, from the above,
We can conclude that the solution for the given system of linear equations is: (-2, 7)

Lesson 9.6 Solving Nonlinear Systems of Equations

Essential Question: How can you solve a system of two equations when one is linear and the other is quadratic?
Answer:
The steps to solve a system of two equations which is linear and quadratic are:
Step 1:
Isolate one of the two variables in one of the equations
Step 2:
Substitute the expression that is equal to the isolated variable from step 1 into the other equation
Step 3:
Solve the resulting quadratic equation to find the x value(s) of the solution(s)

EXPLORATION 1
Solving a System of Equations
Work with a partner. Solve the system of equations by graphing each equation and finding the points of intersection.
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 1
System of Equations
y = x + 2
y = x2 + 2x
Answer:
The given system of equations are:
y = x + 2
y = x² + 2x
So,
The representation of the given system of equations in the coordinate plane is:

So,
From the graph,
We can observe the intersection point of the 2 equations
So,
The intersection point of the given system of equations is: (1, 3)
Hence, from the above,
We can conclude that the solution of the given system of equations is: (1, 3)

EXPLORATION 2

Analyzing Systems of Equations
Work with a partner. Match each system of equations with its graph. Then solve the system of equations.
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 2
Answer:
a.
The given system of equations are:
y = x² – 4
y = -x – 2
So,
x² – 4 = -x – 2
(x – 2) (x + 2) = -(x + 2)
x – 2 = -1
x = -1 + 2
x = 1
So,
For x = 1,
y = -1 – 2
y = -3
So,
The solution or intersection point of the given system of equations is: (1, -3)
Hence, from the above graphs,
We can conclude that graph A matches with the solution of the given system of equations

b.
The given system of equations are:
y = x² – 2x + 2
y = 2x – 2
So,
x² – 2x + 2 = 2x – 2
x² – 2x – 2x + 2 + 2 = 0
x² – 4x + 4 = 0
(x – 2)² = 0
√x – 2 = √0
x – 2 = 0
x = 2
Now,
For x = 2,
y = 2 (2) – 2
= 4 – 2
= 2
So,
The solution or intersection point of the given system of equations is: (2, 2)
Hence, from the above graphs,
We can conclude that graph C matches with the solution of the given system of equations

c.
The given system of equations are:
y = x² + 1
y = x – 1
So,
x² + 1 = x – 1
x² – x + 1 + 1 = 0
x²- x + 2 = 0
Now,
From the above equation,
We can observe that we can’t write the equation in factor form
So,
There will no solution or intersection point for the given system of equations
Hence, from the above graphs,
We can conclude that graph B matches with the solution of the given system of equations

d.
The given system of equations are:
y = x² – x – 6
y = 2x – 2
So,
x² – x – 6 = 2x – 2
x² – x – 2x – 6 + 2= 0
x² – 3x – 4 = 0
x² – 4x + x – 4 = 0
x (x – 4) + 1 ( x – 4) = 0
(x + 1) (x – 4) = 0
x + 1 = 0 or x – 4 = 0
x = -1 or x = 4
Now,
For x = -1,
y = 2 (-1) – 2
= -2 – 2
= -4
For x = 4,
y = 2 (4) – 2
= 8 – 2
= 6
So,
The solutions or intersection points of the given system of equations are: (-1, -4), and (4, 6)
Hence, from the above graphs,
We can conclude that graph D matches with the solution of the given system of equations

Communicate Your Answer

Question 3.
How can you solve a system of two equations when one is linear and the other is quadratic?
Answer:
The steps to solve a system of two equations which is linear and quadratic are:
Step 1:
Isolate one of the two variables in one of the equations
Step 2:
Substitute the expression that is equal to the isolated variable from step 1 into the other equation
Step 3:
Solve the resulting quadratic equation to find the x value(s) of the solution(s)

Question 4.
Write a system of equations (one linear and one quadratic) that has (a) no solutions, (b) one solution, and (c) two solutions. Your systems should be different from those in Explorations 1 and 2.
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 3
Answer:
The system of equations (one linear and one quadratic) that has
a) No solutions:
y = x² – 5x + 7
y = x – 3

b) One solution:
y = x² – 12x + 36
y = x + 7

c) Two solutions:
y = 3x² + 11x + 6
y = x + 10

Monitoring Progress

Solve the system by graphing.
Question 1.
y = x2 + 4x – 4
y = 2x – 5
Answer:
The given system of equations are:
y = x² + 4x – 4
y = 2x – 5
So,
The representation of the given system of equations in the coordinate plane is:

So,
From the graph,
We can observe the intersection point of the 2 equations
So,
The intersection point of the given system of equations is: (-6,-6)
Hence, from the above,
We can conclude that the solution of the given system of equations is: (-6,-6)

Question 2.
y = -x + 6
y = -2x2 – x + 3
Answer:
The given system of equations are:
y = -x + 6
y = -2x² – x + 3
So,
The representation of the given system of equations in the coordinate plane is:

So,
From the graph,
We can observe that there is no intersection point of the 2 equations
Hence, from the above,
We can conclude that there is no solution for the given system of equations

Question 3.
y = 3x – 15
y = \(\frac{1}{2}\)x2 – 2x – 7
Answer:
The given system of equations are:
y = 3x – 15
y = \(\frac{1}{2}\)x2 – 2x – 7
So,
The representation of the given system of equations in the coordinate plane is:

So,
From the graph,
We can observe the intersection points of the 2 equations
So,
The intersection points of the given system of equations are: (8, 9) and (2, -9)
Hence, from the above,
We can conclude that the solutions of the given system of equations are: (8, 9) and (2, -9)

Solve the system by substitution.
Question 4.
y = x2 + 9 5.
y = 9
Answer:
The given system of equations are:
y = x² + 9.5
y = 9
So,
x² + 9.5 = 9
x² = 9 – 9.5
x² = -0.5
We know that,
The square of a real nuber can’t be negative
Hence, from the above,
We can conclude that the given system of equations has no real solutions

Question 5.
y = -5x
y = x2 – 3x – 3
Answer:
The given system of equations are:
y = -5x
y = x² – 3x – 3
So,
-5x = x² – 3x – 3
x² – 3x + 5x – 3 = 0
x² + 2x – 3 = 0
x² + 3x – x – 3 = 0
x (x + 3) – 1 ( x+ 3) = 0
(x – 1) (x + 3) = 0
x – 1 = 0 or x + 3 = 0
x = 1 or x = -3
Now,
For x = 1,
y = -5 (1)
= -5
For x = -3,
y = -5(-3)
= 15
So,
The intersection points of the given system of equations are: (1, -5) and (-3, 15)
Hence, from the above,
We can conclude that the solutions for the given system of equations are: (1, -5) and (-3, 15)

Question 6.
y = -3x2 + 2x + 1
y = 5 – 3x
Answer:
The given system of equations are:
y = -3x² + 2x + 1
y = 5 – 3x
So,
-3x² + 2x + 1 = 5 – 3x
-3x² + 2x + 3x + 1 – 5 = 0
-3x² + 5x – 4 = 0
3x² – 5x + 4 = 0
Compare the given equation with
ax² + bx + c = 0
So,
a = 3, b = -5, and c = 4
So,
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{5 ± \sqrt{(-5)² – 4 (3) (4)}}{2(3)}\)
x = \(\frac{5 ± \sqrt{-23}}{6}\)
We know that,
The square of a real nuber can’t be negative
Hence, from the above,
We can conclude that the given system of equations has no real solutions

Solve the system by elimination.
Question 7.
y = x2 + x
y = x + 5
Answer:
The given system of equations are:
y = x² + x —–(1)
y = x + 5 ——(2)
From the above equations,
We can observe that the variable n the left side of both the equations are equal
So,
eq (1) – eq (2)
So,
y – y = x² + x – x – 5
0 = x² – 5
x² = 5
√x² = √5
x = ±√5
So,
For x = ±√5,
y = ±√5 + 5
Hence, from the above,
We can conclude that the solutions for the given system of equations are: (±√5, ±√5 + 5)

Question 8.
y = 9x2 + 8x – 6
y = 5x – 4
Answer:
The given system of equations are:
y = 9x² + 8x – 6 —–(1)
y = 5x – 6 —–(2)
From the above equations,
We can observe that the variable on the left side of the equations are equal
So,
eq (1) – eq (2)
y – y = 9x² + 8x – 6 – 5x + 4
0 = 9x²+ 3x – 2
9x² + 3x – 2 = 0
9x² + 6x – 3x – 2 = 0
3x(3x – 1) + 2 (3x – 1) = 0
(3x + 2) (3x – 1) = 0
3x + 2 = 0 or 3x – 1 = 0
3x = -2 or 3x = 1
x = –\(\frac{2}{3}\) or x = \(\frac{1}{3}\)
Now,
For x = –\(\frac{2}{3}\),
y = 5 (-\(\frac{2}{3}\)) – 4
y = –\(\frac{22}{3}\)
For x = \(\frac{1}{3}\),
x = 5 (\(\frac{1}{3}\)) – 4
x = –\(\frac{7}{3}\)
Hence, from the above,
We can conclude that the solutions for the given system of equations are: (-\(\frac{2}{3}\), –\(\frac{22}{3}\)), and (\(\frac{1}{3}\), –\(\frac{7}{3}\))

Question 9.
y = 2x + 5
y = -3x2 + x – 4
Answer:
The given system of equations are:
y = 2x + 5 —-(1)
y = -3x² + x – 4 —–(2)
From the above equations,
We can observe that the variable on the left side of the equations are equal
So,
eq (1) – eq (2)
y – y = 2x + 5 + 3x² – x + 4
0 = 3x² + x + 9
3x² + x + 9 = 0
Now,
Compare the above equation with
ax² + bx + c = 0
So,
a = 3, b = 1, and c = 9
Now,
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{-1 ± \sqrt{(1)² – 4 (3) (9)}}{2(3)}\)
x = \(\frac{-1 ± \sqrt{-107}}{6}\)
We know that,
The square of a real number can not be negative
Hence, from the above,
We can conclude that the given system of equations has no real solutions

Use the method in Example 4 to approximate the solution(s) of the system to the nearest thousandth.
Question 10.
y = 4x
y = x2 + x + 3
Answer:
The given system of equations are:
y = 4^x
y = x² + x + 3
So,
The representation of the given system of equations in the coordinate plane is:

Hence, from the above,
We can conclude that the solution of the given system of equations lies between -2 and -1

Question 11.
y = 4x2 – 1
y = -2(3)x + 4
Answer:
The given system of equations are:
y = 4x² – 1
y = -2 (3^x) + 4
So,
The representation of the given system of equations in the coordinate plane is:

Hence, from the above,
We can conclude that the solution for the given system of equations lies between 0 and 1

Question 12.
y = x2 + 3x
y = -x2 + x + 10
Answer:
The given system of equations are:
y = x² + 3x
y = -x² + x + 10
So,
The representation of the given system of equations in the coordinate plane is:

Hence, from the above,
We can conclude that the solution for the given system of equations lies between -3 and -2.5

Solve the equation. Round your solution(s) to the nearest hundredth.
Question 13.
3x – 1 = x2 – 2x + 5
Answer:
The given equation is:
3^x – 1 = x² – 2x + 5
So,
The representation of the given equation in the coordinate plane is:

Hence, from the above,
We can conclude that the solution for the given system of equations lies between 1 and 1.5

Question 14.
4x2 + x = -2 (\(\frac{1}{2}\))x + 5
Answer:
The given equation is:
4x2 + x = -2 (\(\frac{1}{2}\))x + 5
So,
The representation of the given equation in the coordinate plane is:

Hence, from the above,
We can conclude that the solution for the given system of equations is: (1, 4)

Solving Nonlinear Systems of Equations 9.6 Exercises

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
Describe how to use substitution to solve a system of nonlinear equations.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 1

Question 2.
WRITING
How is solving a system of nonlinear equations similar to solving a system of linear equations? How is it different?
Answer:
It is similar because you can use the same methods for linear and non-linear equations
However, the difference is that sometimes you also need to use different methods and you can obtain multiple solutions

In Exercises 3–6, match the system of equations with its graph. Then solve the system.
Question 3.
y = x2 – 2x + 1
y = x + 1
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 3

Question 4.
y = x2 + 3x + 2
y = -x – 3
Answer:
The given system of equations are:
y = x² + 3x + 2
y = -x – 3
So,
x² + 3x + 2 = -x – 3
x² + 3x + 2 + x + 3 = 0
x²+ 4x + 5 = 0
Now,
From the above equation,
We can observe that we can’t write the equation in factor form
So,
There will no solution or intersection point for the given system of equations
Hence, from the above graphs,
We can conclude that graph D matches with the solution of the given system of equations

Question 5.
y = x – 1
y = -x2 + x – 1
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 5

Question 6.
y = -x + 3
y = -x2 – 2x + 5
Answer:
The given system of equations are:
y = -x + 3
y = -x² – 2x + 5
So,
-x + 3 = -x² – 2x + 5
x² + 2x – 5 – x + 3 = 0
x² + x – 2 = 0
x² + 2x – x – 2 = 0
x (x + 2) – 1 (x + 2) = 0
(x – 1) (x + 2) = 0
x – 1 = 0 or x + 2 = 0
x = 1 or x = -2
Now,
For x = 1,
y = -1 + 3
= 2
For x = -2,
y = 2 + 3
= 5
So,
The solutions or the intersection points of the given system of equations are: (1, 2), and (-2, 5)
Hence, from the above,
We can conclude that the graph C matches with the given system of equations

Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 4

In Exercises 7–12, solve the system by graphing.
Question 7.
y = 3x2 – 2x + 1
y = x + 7
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 7

Question 8.
y = x2 + 2x + 5
y = -2x – 5
Answer:
The given system of equations are:
y = x² + 2x + 5
y = -2x – 5
So,
The representation of the given system of equations in the coordinate plane is:

So,
From the graph,
We can observe that there are no intersection points in the graph,
Hence, from the above,
We can conclude that there are no solutions for the given system of equations

Question 9.
y = -2x2 – 4x
y = 2
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 9

Question 10.
y = \(\frac{1}{2}\)x2 – 3x + 4
y = x – 2
Answer:
The given system of equations are:
y = \(\frac{1}{2}\)x² – 3x + 4
y = x – 2
So,
The representation of the given system of equations in the coordinate plane is:

So,
From the graph,
We can observe that the solution or intersection point of the given system of equations is: (6, 4)
Hence, from the above,
We can conclude that the solution of the given system of equations is: (6, 4)

Question 11.
y = \(\frac{1}{3}\) x2 + 2x – 3
y = 2x
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 11.1
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 11.2

Question 12.
y = 4x2 + 5x – 7
y = -3x + 5
Answer:
The given system of equations are:
y = 4x² + 5x – 7
y = -3x + 5
So,
The representation of the given system of equations in the coordinate plane is:

So,
From the graph,
We can observe that
The solution or the intersection point of the given system of equations is: (1, 2)
Hence, from the above,
We can conclude that the solution of the given system of equations is: (1, 2)

In Exercises 13–18, solve the system by substitution.
Question 13.
y = x – 5
y = x2 + 4x – 5
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 13

Question 14.
y = -3x2
y = 6x + 3
Answer:
The give system of equations are:
y = -3x²
y = 6x + 3
So,
-3x² = 6x + 3
3x² + 6x + 3 = 0
3x² + 3x + 3x + 3 = 0
3x (x + 1) + 3 (x + 1) = 0
(3x + 3) (x + 1) = 0
3x + 3 = 0 or x + 1 = 0
3x = -3 or x = -1
x = -1 or x = -1
So,
For x = -1,
y = 6 (-1) + 3
= -6 + 3
= -3
Hence, from the above,
We can conclude that the solution of the given system of equations is: (-1, -3)

Question 15.
y = -x + 7
y = -x2 – 2x – 1
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 15

Question 16.
y = -x2 + 7
y = 2x + 4
Answer:
The given system of equations are:
y = -x² + 7
y = 2x + 4
So,
-x² + 7 = 2x + 4
x² – 7 + 2x + 4 = 0
x² + 2x – 3 = 0
x² + 3x – x – 3 = 0
x (x + 3) – 1 (x + 3) = 0
(x – 1) (x + 3) = 0
x – 1 = 0 or x + 3 = 0
x = 1 or x = -3
Now,
For x = 1,
y = 2 (1) + 4
= 6
For x = -3,
y = 2 (-3) + 4
= -2
Hence, from the above,
We can conclude that the solutions for the given system of equations are: (1, 6), and (-3, -2)

Question 17.
y – 5 = -x2
y = 5
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 17

Question 18.
y = 2x2 + 3x – 4
y – 4x = 2
Answer:
The given system of equations are:
y = 2x² + 3x – 4
y – 4x = 2
So,
y = 4x + 2
Now,
2x² + 3x – 4 = 4x + 2
2x² + 3x – 4x – 4 – 2 = 0
2x² – x – 6 = 0
2x² – 4x + 3x – 6 = 0
2x (x – 2) + 3 (x – 2) = 0
(2x + 3) (x – 2) = 0
2x + 3 = 0 or x – 2 = 0
2x = -3 or x = 2
x = –\(\frac{3}{2}\) or x = 2
Now,
For x = –\(\frac{3}{2}\),
y = 4 (-\(\frac{3}{2}\)) + 2
= -4
For x = 2,
y = 4 (2) + 2
= 10
Hence, from the above,
We can conclude that the solutions of the given system of equations are: (-\(\frac{3}{2}\), -4), and (2, 10)

In Exercises 19–26, solve the system by elimination.
Question 19.
y = x2 – 5x – 7
y = -5x + 9
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 19

Question 20.
y = -3x2 + x + 2
y = x + 4
Answer:
The given system of equations are:
y = -3x² + x + 2 —- (1)
y = x + 4 —- (2)
Since the variables on the left side of the 2 equations are equal,
eq (1) – eq (2)
So,
-3x² + x + 2 – x – 4 = 0
-3x² + x – x = 4 – 2
-3x² = 2
3x² = -2
We know that,
The square of a real number will not take any negative values
Hence, from the above,
We can conclude that the given system of equations oes not have any real solutions

Question 21.
y = -x2 – 2x + 2
y = 4x + 2
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 21

Question 22.
y = -2x2 + x – 3
y = 2x – 2
Answer:
The given system of equations are:
y = -2x² + x – 3 —-(1)
y = 2x – 2 —-(2)
From the above equations,
We can observe that the variable on the left side of the given system of equations is equal
So,
eq (1) – eq (2)
So,
-2x² + x – 3 – 2x + 2 = 0
-2x² – x – 1 = 0
2x² + x + 1 = 0
Compare the above equation with
ax² + bx + c = 0
So,
a = 2, b = 1, and c = 1
Now,
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{-1 ± \sqrt{1² – 4 (2) (1)}}{2 (2)}\)
x = \(\frac{-1 ± \sqrt{-7}}{4}\)
We know that,
The square of a real number will not take any negative values
Hence, from the above,
We can conclude that the given system of equations does not have any real solutions

Question 23.
y = 2x – 1
y = x2
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 23

Question 24.
y = x2 + x + 1
y = -x – 2
Answer:
The given system of equations are:
y = x² + x + 1 —-(1)
y = -x – 2 —-(2)
From the above equations,
We can observe that the variable on the left side of the given system of equations is equal,
eq (1) – eq (2)
So,
x² + x + 1 + x + 2 = 0
x² + 2x + 3 = 0
Compare the above equation with
ax² + bx + c = 0
So,
a = 1, b = 2, and c = 3
Now,
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{-2 ± \sqrt{2² – 4 (3) (1)}}{2 (1)}\)
x = \(\frac{-2 ± \sqrt{-8}}{2}\)
We know that,
The square of a real number will not take any negative values
Hence, from the above,
We can conclude that the given system of equations does not have any real solutions

Question 25.
y + 2x = 0
y = x2 + 4x – 6
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 25.1
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 25.2

Question 26.
y = 2x – 7
y + 5x = x2 – 2
Answer:
The given system of equations are:
y = 2x – 7
y + 5x = x²
So,
y = 2x – 7 —– (1)
y = x² – 5x —–(2)
From the given system of equations,
e can observe that the variable on the left side of the given system of equations is equal
So,
eq (1) – eq (2)
So,
2x – 7 – x² + 5x = 0
-x² + 7x – 7 = 0
x² – 7x + 7 = 0
Compare the above equation with
ax² + bx + c = 0
So,
a = 1, b = -7, and c = 7
Now,
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{7 ± \sqrt{(-7)² – 4 (7) (1)}}{2 (1)}\)
x = \(\frac{7 ± \sqrt{21}}{2}\)
x = \(\frac{7 + \sqrt{21}}{2}\) or x = \(\frac{7 – \sqrt{21}}{2}\)
x = 5.79 or x = 1.20
Hence, from the above,
We can conclude that the solutions for the given system of equations are: 5.79 and 1.20

Question 27.
ERROR ANALYSIS
Describe and correct the error in solving the system of equations by graphing.
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 5
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 27

Question 28.
ERROR ANALYSIS
Describe and correct the error in solving for one of the variables in the system.
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 6
Answer:
The given system of equations are:
y = 3x² – 6x + 4
y = 4
So,
3x² – 6x + 4 = 4
3x²- 6x = 4 – 4
3x²= 6x
3x = 6
x = \(\frac{6}{3}\)
x = 2
Hence, from the above,
We can conclude that the solution of the given system of equations is: (2, 4)

In Exercises 29–32, use the table to describe the locations of the zeros of the quadratic function f.
Question 29.
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 7
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 29

Question 30.
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 8
Answer:
The given table is:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 8
From the above table,
We can observe that
There is a sign change between x = 1 and x = 2 and f (1) and f (2) are in the same distance,
The first solution lies between 1 and 2
There is a sign change between x = 3 and x = 4 and f (3) and f (4) are in the same distance,
The second solution lies between 3 and 4
Hence, from the above,
We can conclude that the solutions for the given function will lie between (1, 2) and (3, 4)

Question 31.
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 9
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 31

Question 32.
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 10
Answer:
The given table is:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 10
From the above table,
We can observe that
There is a sign change between x = 2 and x = 3 and f (2) and f (3) are in the same distance,
But, f (3) is very close to 0
So,
The first solution is closer to 3 than 2
There is a sign change between x = 5 and x = 6 and f (5) and f (6) are in the same distance,
But, f (5) is very close to 0
So,
The second solution is closer to 5 than 6
Hence, from the above,
We can conclude that the solution for the given function will lie between (3, 5)

In Exercises 33–38, use the method in Example 4 to approximate the solution(s) of the system to the nearest thousandth.
Question 33.
y = x2 + 2x + 3
y = 3x
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 33.1
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 33.2

Question 34.
y = 2x + 5
y = x2 – 3x + 1
Answer:
The given system of equations are:
y = 2^x + 5
y = x² – 3x + 1
So,
The representation of the given system of equations in the coordinate plane is:

Hence, from the above,
We can conclude that the solution for the given system of equations is approximately (1, 6)

Question 35.
y = 2(4)x – 1
y = 3x2 + 8x
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 35.1
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 35.2

Question 36.
y = -x2 – 4x – 4
y = -5x – 2
Answer:
The given system of equations are:
y = -x² – 4x – 4
y = -5^x – 2
So,
The representation of the given system of equations in the coordinate plane is:

Hence, from the above,
We can conclude that the solution for the given system of equations is approximately (-1, -3)

Question 37.
y = -x2 – x + 5
y = 2x2 + 6x – 3
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 37.1
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 37.2

Question 38.
y = 2x2 + x – 8
y = x2 – 5
Answer:
The given system of equations are:
y = 2x² + x – 8
y = x² – 5
So,
The representation of the given system of equations in the coordinate plane is:

Hence, from the above,
We can conclude that the approximate solution for the given system of equations is: (-1, 2)

In Exercises 39–46, solve the equation. Round your solution(s) to the nearest hundredth.
Question 39.
3x + 1 = x2 + 7x – 1
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 39

Question 40.
-x2 + 2x = -2x + 5
Answer:
The given equation is:
-x² + 2x = -2x + 5
So,
The representation of the given equation in the coordinate palne is:

Hence, from the above,
We can conclude that there are no solutions for the given system of equations

Question 41.
x2 – 6x + 4 = -x2 – 2x
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 41

Question 42.
2x2 + 8x + 10 = -x2 – 2x + 5
Answer:
The given equation is:
2x² + 8x + 10 = -x² – 2x + 5
So,
2x² + x² + 8x + 2x + 10 – 5 = 0
3x² + 10x + 5 = 0
So,
The representation of the given equation in the coordinate plane is:

Hence, from the above,
We can conclude that the approximate solution for the given equation is (-1, 6)

Question 43.
-4 (\(\frac{1}{2}\))x = -x2 – 5
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 43

Question 44.
1.5(2)x – 3 = -x2 + 4x
Answer:
The given equation is:
1.5(2)^x – 3 = -x² + 4x
So,
The representation of the given equation in the coordiante plane is:

Hence, from the above,
We can conclude that the approximate solutions for the given equation are: (2, 4) and ((0, -2)

Question 45.
8x-2 + 3 = 2 (\(\frac{3}{2}\))x
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 45

Question 46.
-0.5(4)x = 5x – 6
Answer:
The given equation is:
-0.5(4)^x = 5^x – 6
So,
The representation of the given equation in the coordinate plane is:

Hence, from the above,
We can conclude that the solution for the given equation is: (1, -2)

Question 47.
COMPARING METHODS
Solve the system in Exercise 37 using substitution. Compare the exact solutions to the approximated solutions.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 47.1
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 47.2

Question 48.
COMPARING METHODS
Solve the system in Exercise 38 using elimination. Compare the exact solutions to the approximated solutions.
Answer:
The given system of equations in Exercise 38 is:
y = 2x² + x – 8 —-(1)
y = x² – 5 —–(2)
Since the variable on the left side in the given system of equations is equal,
eq (1) – eq(2)
2x² + x – 8 = x² – 5
2x² – x² + x – 8 + 5 = 0
x² + x – 3 = 0
Now,
Compare the above equation with
ax² + bx + c = 0
So,
a = 1, b= 1, and c = -3
So,
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{-1 ± \sqrt{1² + 4 (1) (3)}}{2 (1)}\)
x = \(\frac{-1 ± \sqrt{13}}{2}\)
x = \(\frac{-1 + \sqrt{13}}{2}\) or x = \(\frac{-1 – \sqrt{13}}{2}\)
We know that,
\(\sqrt{13}\) = 3.60
So,
x = 1.30 or x = -2.30
Hence, from the above,
We can conclude that the exact solutions and approximate solutions are the same

Question 49.
MODELING WITH MATHEMATICS
The attendances y for two movies can be modeled by the following equations, where x is the number of days since the movies opened.
y = -x2 + 35x + 100 Movie A
y = -5x + 275 Movie B
When is the attendance for each movie the same?
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 49

Question 50.
MODELING WITH MATHEMATICS
You and a friend are driving boats on the same lake. Your path can be modeled by the equation y = -x2 – 4x – 1, and your friend’s path can be modeled by the equation y = -2x + 8. Do your paths cross each other? If so, what are the coordinates of the point(s) where the paths meet?
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 11
Answer:
The given system of equations is:
y = -x² – 4x – 1 ——-> Your path
y = -2x + 8 ———–> Your friend’s path
So,
The representation of the paths of yours and your friend in the coordinate plane is:
Hence, from the above graph,
We can conclude that the paths of yours and your friend did not cross each other

Question 51.
MODELING WITH MATHEMATICS
The arch of a bridge can be modeled by y = -0.002x2 + 1.06x, where x is the distance (in meters) from the left pylons and y is the height (in meters) of the arch above the water. The road can be modeled by the equation y = 52. To the nearest meter, how far from the left pylons are the two points where the road intersects the arch of the bridge?
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 12
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 51

Question 52.
MAKING AN ARGUMENT
Your friend says that a system of equations consisting of a linear equation and a quadratic equation can have zero, one, two, or infinitely many solutions. Is your friend correct? Explain.
Answer:

COMPARING METHODS In Exercises 53 and 54, solve the system of equations by (a) graphing, (b) substitution, and (c) elimination. Which method do you prefer? Explain your reasoning.
Question 53.
y = 4x + 3
y = x2 + 4x – 1
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 53.1
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 53.2

Question 54.
y = x2 – 5
y = -x + 7
Answer:
The given system of equations re:
y = x² – 5 —— (1)
y = -x + 7 —— (2)
a) By graphing:
The representation of the given system of equations in the coordinate plane is:

Hence, from the above,
We can conclude that the solution for the given system of equations is: (3, 4)
b) By substitution:
Since the variable on the left side of the given system of equations is equal,
x² – 5 = -x + 7
x² + x – 5 – 7 = 0
x² + x – 12 = 0
x² + 4x – 3x – 12 = 0
x (x – 3) + 4 (x – 3) = 0
(x + 4) (x – 3) = 0
x + 4 = 0 or x – 3 = 0
x = -4 or x = 3
So,
For x = 3,
y = -3 + 7
= 4
For x = -4,
y = 4 + 7
= 11
Hence, from the above,
We can conclude that the solutions for the given system of equations are: (3, 4) and (-4, 11)
c) By elimination:
Since the variable on the left side of the given system of equations is equal,
eq (1) – eq (2) = 0
So,
x² + x – 5 – 7 = 0
x² + x – 12 = 0
x² + 4x – 3x – 12 = 0
x (x – 3) + 4 (x – 3) = 0
(x + 4) (x – 3) = 0
x + 4 = 0 or x – 3 = 0
x = -4 or x = 3
So,
For x = 3,
y = -3 + 7
= 4
For x = -4,
y = 4 + 7
= 11
Hence, from the above,
We can conclude that the solutions for the given system of equations are: (3, 4) and (-4, 11)
Hence, from all the above three methods,
We can conclude that  we can prefer the substitution method because of the factorization method

Question 55.
MODELING WITH MATHEMATICS
The function y = -x2 + 65x + 256 models the number y of subscribers to a website, where x is the number of days since the website launched. The number of subscribers to a competitor’s website can be modeled by a linear function. The websites have the same number of subscribers on Days 1 and 34.
a. Write a linear function that models the number of subscribers to the competitor’s website.
b. Solve the system to verify the function from part (a).
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 55.1
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 55.2

Question 56.
HOW DO YOU SEE IT?
The diagram shows the graphs of two equations in a system that has one solution.
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 13
a. How many solutions will the system have when you change the linear equation to y = c + 2?
Answer:
From the given graph,
We can observe that y = c is a constant line that is parallel to the x-axis
So,
y = c + 2 is also a constant line
Hence, from the above,
We can conclude that the system will have only one solution when you change the linear equation to y = c + 2

b. How many solutions will the system have when you change the linear equation to y = c – 2?
Answer:
From the given graph,
We can observe that y = c is a constant line that is parallel to the x-axis
So,
y = c – 2 is also a constant line
Hence, from the above,
We can conclude that the system will have only one solution when you change the linear equation to y = c – 2

Question 57.
WRITING
A system of equations consists of a quadratic equation whose graph opens up and a quadratic equation whose graph opens down. Describe the possible number of solutions of the system. Sketch examples to justify your answer.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 57

Question 58.
PROBLEM-SOLVING
The population of a country is 2 million people and increases by 3% each year. The country’s food supply is sufficient to feed 3 million people and increases at a constant rate that feeds 0.25 million additional people each year.
a. When will the country first experience a food shortage?
Answer:

b. The country doubles the rate at which its food supply increases. Will food shortages still occur? If so, in what year?
Answer:
In part (b), the population of the country will not change
Only the rate will change
So,

Question 59.
ANALYZING GRAPHS
Use the graphs of the linear and quadratic functions.
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 14
a. Find the coordinates of point A.
b. Find the coordinates of point B.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 59.1
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 59.2

Question 60.
THOUGHT-PROVOKING
Is it possible for a system of two quadratic equations to have exactly three solutions? exactly four solutions? Explain your reasoning. (Hint: Rotations of the graphs of quadratic equations still represent quadratic equations.)
Answer:
We know that,
The quadratic equation is in the form of
ax² + bx + c = 0
So,
From the above form,
We can say that the quadratic equation will have only an even number of solutions even we consider the rotations of the graphs of the quadratic equations
Hence, from the above,
We can conclude that the quadratic equation will have exactly four solutions

Question 61.
PROBLEM-SOLVING
Solve the system of three equations shown.
y = 2x – 8
y = x2 – 4x – 3
y = -3(2)x
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 61

Question 62.
PROBLEM-SOLVING
Find the point(s) of intersection, if any, of the line y = -x – 1 and the circle x2 + y2 = 41.
Answer:
The given equations are:
y = -x – 1 —-  (1)
x² + y² = 41 —- (2)
So,
Substitute eq (1) into eq (2)
x² + (-x – 1)² = 41
x² + x² + 1 + 2x = 41
2x² + 2x – 40 = 0
x²+ x – 20 = 0
x² + 5x – 4x – 20 = 0
x ( x + 5 ) – 4 (x + 5) = 0
(x – 4) (x + 5) = 0
x – 4 = 0 or x + 5 = 0
x = 4 or x = -5
So,
For x = 4,
y = -4 – 1
= -5
For x = -5,
y = 5 – 1,
= 4
Hence, from the above,
We acn conclude that the points of intersection for the given equations are: (4, -5) and (-5, 4)

Maintaining Mathematical Proficiency

Graph the system of linear inequalities.
Question 63.
y > 2x
y > -x + 4
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 63

Question 64.
y ≥ 4x + 1
y ≤ 7
Answer:
The given pair of inequalities are:
y ≥ 4x + 1
y ≤ 7
Hence,
The representtaion of the given pair of inequalities in the coordinate plane is:

Question 65.
y – 3 ≤ -2x
y + 5 < 3x
Answer:
The given pair of inequalities are:
y – 3≤ -2x
y + 5 < 3x
Hence,
The representation of the given pair of inequalities in the coordinate plane is:

Question 66.
x + y > -6
2y ≤ 3x + 4
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 65

Graph the function. Describe the domain and range.
Question 67.
y = 3x2 + 2
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 67

Question 68.
y = -x2 – 6x
Answer:
The given equation is:
y = -x² – 6x
So,
The representation of the given equation in the coordinate plane is:

We know that,
The domain is the range of values of x that make the equation positive
The range is the range of values of y that make the equation positive
Hence, from the above,
We can conclude that
The domain of the given function is: 0 ≤ x ≤ -6
The range of the given function is: -10 ≤ y ≤ 9

Question 69.
y = -2x2 + 12x – 7
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 a 69

Question 70.
y = 5x2 + 10x – 3
Answer:
The given equation is:
y = 5x² + 10x – 3
So,
The representation of the given equation in the coordinate plane is:

We know that,
The domain is the set of all the values of x that make the equation positive
The range is the set of all the values of y that make the equation positive
Hence, from the above,
We can conclude that
The approximate domain of the given function is: -2 ≤ x ≤ 0
The range of the given equation is: -8 ≤ y ≤ 10

Solving Quadratic Equations Performance Task: Form Matters

9.4–9.6 What Did You Learn?

Core Vocabulary
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 15

Core Concepts
Section 9.4
Completing the Square, p. 506

Section 9.5
Quadratic Formula, p. 516
Interpreting the Discriminant, p. 518

Section 9.6
Solving Systems of Nonlinear Equations, p. 526

Mathematical Practices
Question 1.
How does your answer to Exercise 74 on page 514 help create a shortcut when solving some quadratic equations by completing the square?
Answer:
We know that,
For the equation that is of the form
ax² + bx + c
When we use the “Completing the squares” method,
We know that,
c = (\(\frac{b}{2}\))² and in place of b,
We will use
b = \(\frac{b}{2}\)

Question 2.
What logical progression led you to your answer in Exercise 55 on page 522?
Answer:
In Exercise 55 on page 522,
We use the sign changes for a and c and find the discriminant to find the logical procession

Question 3.
Compare the methods used to solve Exercise 53 on page 532. Discuss the similarities and differences among the methods.
Answer:

Performance Task Form Matters

Each form of a quadratic function has its pros and cons. Using one form, you can easily find the vertex, but the zeros are more difficult to find. Using another form, you can easily find the y-intercept, but the vertex is more difficult to find. Which form would you use in different situations? How can you convert one form into another? To explore the answers to these questions and more, go to.
Big Ideas Math Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations 9.6 16

Solving Quadratic Equations Chapter Review

9.1 Properties of Radicals (pp. 479–488)

Simplify the expression.
Question 1.
\(\sqrt{72 p^{7}}\)
Answer:
The given expression is:
\(\sqrt{72 p^{7}}\)
= \(\sqrt{72 p^{6} p^{1}}\)
= \(\sqrt{36 p^{6} × 2p^{1}}\)
= 6p³ \(\sqrt{2p}\)
Hence, from the above,
We can conclude that
\(\sqrt{72 p^{7}}\) = 6p³ \(\sqrt{2p}\)

Question 2.
\(\sqrt{\frac{45}{7 y}}\)
Answer:
The given expression is:
\(\sqrt{\frac{45}{7 y}}\)
= \(\frac{\sqrt{45}}{\sqrt{7 y}}\)
= \(\frac{\sqrt{5 (9)}}{\sqrt{7 y}}\)
= 3\(\sqrt{\frac{5}{7 y}}\)
Hence, from the above,
We can conclude that
\(\sqrt{\frac{45}{7 y}}\) = 3\(\sqrt{\frac{5}{7 y}}\)

Question 3.
\(\sqrt[3]{\frac{125 x^{11}}{4}}\)
Answer:
The given expression is:
\(\sqrt[3]{\frac{125 x^{11}}{4}}\)
= \(\sqrt[3]{\frac{5 ^{3} x^{9} x^{2}}{4}}\)
= 5x³\(\sqrt[3]{\frac{ x^{2}}{4}}\)
Hence, from the above,
We can conclude that
\(\sqrt[3]{\frac{125 x^{11}}{4}}\) = 5x³\(\sqrt[3]{\frac{ x^{2}}{4}}\)

Question 4.
\(\frac{8}{\sqrt{6}+2}\)
Answer:
The given expression is:
\(\frac{8}{\sqrt{6}+2}\)
To make the given expression rational,
Multiply and divide the given expression with \(\sqrt{6} – 2\)
So,
\(\frac{8}{\sqrt{6}+2}\)
= \(\frac{8}{\sqrt{6}+2}\) × \(\frac{\sqrt{6} – 2}{\sqrt{6} – 2}\)
= \(\frac{8 (\sqrt{6} – 2} {6 – 4}\)
= \(\frac{8\sqrt{6} – 2}{2}\)
= 4 \(\sqrt{6} – 2\)
Hence, from the above,
We can conclude that
\(\frac{8}{\sqrt{6}+2}\) = 4 \(\sqrt{6} – 2\)

Question 5.
4\(\sqrt{3}\) + 5\(\sqrt{12}\)
Answer:
The given expression is:
4\(\sqrt{3}\) + 5\(\sqrt{12}\)
= 4\(\sqrt{3}\) + 5\(\sqrt{4 (3)}\)
= 4\(\sqrt{3}\) + 5 (2)\(\sqrt{3}\)
= 4\(\sqrt{3}\) + 10\(\sqrt{3}\)
= \(\sqrt{3}\) (5 + 4)
= 9\(\sqrt{3}\)
Hence, from the above,
We can conclude that
4\(\sqrt{3}\) + 5\(\sqrt{12}\) = 9\(\sqrt{3}\)

Question 6.
15\(\sqrt [ 3 ]{ 2 }\) – 2\(\sqrt [ 3 ]{ 54 }\)
Answer:
The given expression is:
15\(\sqrt [ 3 ]{ 2 }\) – 2\(\sqrt [ 3 ]{ 54 }\)
= 15\(\sqrt [ 3 ]{ 2 }\) – 2\(\sqrt [ 3 ]{ 27 (2) }\)
= 15\(\sqrt [ 3 ]{ 2 }\) – 2 (3)\(\sqrt [ 3 ]{ 2 }\)
= 15\(\sqrt [ 3 ]{ 2 }\) – 6\(\sqrt [ 3 ]{ 2 }\)
= [/latex]\sqrt[3]{2}[/latex] (15 – 6)
= 9 \(\sqrt[3]{2}\)
Hence, from the above,
We can conclude that
15\(\sqrt [ 3 ]{ 2 }\) – 2\(\sqrt [ 3 ]{ 54 }\) = 9\(\sqrt[3]{2}\)

Question 7.
(3\(\sqrt{7}\) + 5)2
Answer:
The given expression is:
(3\(\sqrt{7}\) + 5)2
= (3\(\sqrt{7}\))² + 5² + 2 (3\(\sqrt{7}\)) (5)
= 9 (7) + 25 + 30\(\sqrt{7}\)
= 88 + 30\(\sqrt{7}\)
Hence, from the above,
We can conclude that
(3\(\sqrt{7}\))² = 88 + 30\(\sqrt{7}\)

Question 8.
\(\sqrt{6}\)(\(\sqrt{18}\) + \(\sqrt{8}\))
Answer:
The given expression is:
\(\sqrt{6}\)(\(\sqrt{18}\) + \(\sqrt{8}\))
= \(\sqrt{6}\)(\(\sqrt{9 (2)}\) + \(\sqrt{4 (2)}\))
= \(\sqrt{6}\)(\(3\sqrt{2}\) + \(2\sqrt{2}\))
= \(\sqrt{6}\)((3 + 2) \(\sqrt{2}\))
= \(\sqrt{6}\) (5\(\sqrt{2}\))
= 5 \(\sqrt{12}\)
= 5 \(\sqrt{4 (3)}\)
= 10 \(\sqrt{3}\)
Hence, from the above,
We can conclude that
\(\sqrt{6}\)(\(\sqrt{18}\) + \(\sqrt{8}\)) = 10 \(\sqrt{3}\)

9.2 Solving Quadratic Equations by Graphing (pp. 489–496)

Solve the equation by graphing
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations cr 9
Question 9.
x2 – 9x + 18 = 0
Answer:
The given equation is:
x² – 9x + 18 = 0
So,
The representation of the given equation in the coordinate plane is:

Hence, from the above,
We can conclude that the solutions of the given equation are: 3 and 6

Question 10.
x2 – 2x = -4
Answer:
The given equation is:
x² – 2x = -4
So,
x² – 2x + 4 = 0
So,
The representation of the given equation in the coordinate plane is:

Hence, from the above,
We can conclude that there are no real solutions for the given equation as there are no x – intercepts

Question 11.
-8x – 16 = x2
Answer:
The given equation is:
-8x – 16 = x²
So,
x² + 8x + 16 = 0
So,
The representation of the given equation in the coordinate plane is:

Hence, from the above,
We can conclude that the solution for the given equation is: -4

Question 12.
The graph of f(x) = (x + 1)(x2 + 2x – 3) is shown. Find the zeros of f.
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations cr 9
Answer:
The given equation is:
f (x) = (x + 1) (x² + 2x – 3)
We know that,
To find the zeros of a function,
We have to make
f (x) = 0
So,
(x + 1) (x²  + 2x – 3) = 0
x + 1 = 0 or x² + 2x – 3 = 0
x + 1 = 0 or x² + 3x – x – 3 = 0
x = -1 or x (x + 3) – 1(x + 3) = 0
x = -1 or (x  1) (x + 3) = 0
x = -1 or x – 1 = 0 or x + 3 = 0
x = -1 or x = 1 or x = -3
Hence, from the above,
We can conclude that the zeroes of the given function is: -1, 1, and -3

Question 13.
Graph f(x) = x2 + 2x – 5. Approximate the zeros of f to the nearest tenth.
Answer:
The given function is:
f (x) = x² + 2x – 5
So,
The representation of the given function in the coordinate plane is:

Hence, from the above,
We can conclude that the approximate solution of the given function is: (-4, 2)

9.3 Solving Quadratic Equations Using Square Roots (pp. 497–502)

Solve the equation using square roots. Round your solutions to the nearest hundredth, if necessary.
Question 14.
x2 + 5 = 17
Answer:
The given equation is:
x² + 5 = 17
Subtract with 5 on both sides
So,
x²+ 5 – 5 = 17 – 5
x² = 12
√x² = √12
x = ±√12
We know that,
√12 = 3.46
So,
x = ±3.46
Hence, from the above,
We can conclude that the solutions for the given equation are: 3.46 and -3.46

Question 15.
x2 – 14 = -14
Answer:
The given equation is:
x² – 14 = -14
Add with 14 on both sides
So,
x² – 14 + 14 = -14 + 1 4
x² = 0
√x² = √0
x = 0
Hence, from the above,
We can conclude that the solution for the given equation is: 0

Question 16.
(x + 2)2 = 64
Answer:
The given equation is:
(x + 2)² = 64
So,
√(x + 2)² = √64
x + 2 = ±8
x + 2 = 8 or x + 2 = -8
x = 8 – 2 or x = -8 – 2
x = 6 or x = -10
Hence, from the above,
We can conclude that the solutions for the given equation is: 6 and -10

Question 17.
4x2 + 25 = -75
Answer:
The given equation is:
4x² + 25 = -75
Subtract with 25 on both sides
4x² + 25 – 25 = -75 – 25
4x² = -100
x² = \(\frac{-100}{25}\)
We know that,
The square of a real number will not take any negative number
Hence, from the above,
We can conclude that there are no real solutions for the given equation

Question 18.
(x – 1)2 = 0
Answer:
The given equation is:
(x – 1)² = 0
So,
√(x – 1)² = √0
x – 1 = 0
x = 0 + 1
x = 1
Hence, from the above,
We can conclude that the solution for the given equation is: 1

Question 19.
19 = 30 – 5x2
Answer:
The given equation is:
19 = 30 – 5x²
Subtract with 19 on both sides
So,
19 – 19 = 30 – 19 – 5x²
0 = 11 – 5x²
5x² = 11
x² = \(\frac{11}{5}\)
√x² = √\(\frac{11}{5}\)
x = ±\(\frac{11}{5}\)
We know that,
\(\frac{11}{5}\) = 2.20
So,
x = ±2.20
Hence, from the above,
We can conclude that the solutions for the given equation are: 2.20 and -2.20

9.4 Solving Quadratic Equations by Completing the Square (pp. 505–514)

Solve the equation by completing the square. Round your solutions to the nearest hundredth, if necessary.
Question 20.
x2 + 6x – 40 = 0
Answer:
The given equation is:
x² + 6x – 40 = 0
Add with 40 on both sides
4² + 6x – 40 + 40 = 40 + 0
x² + 6x = 40
Compare the above equation with
ax² + b = d
We know that,
To complete the square,
We have to know that value of c
So,
c = (\(\frac{b}{2}\))²
So,
The completed equation will be in the form
ax² + bx + c = d
So,
x² + 6x + (\(\frac{6}{2}\))² = 40 + (\(\frac{6}{2}\))²
x² + 6x + 9 = 40 + 9
(x + 3)² = 49
√(x + 3)² = √49
x + 3 = 7 or x + 3 = -7
x = 7 – 3 or x = -7 – 3
x = 4 or x = -10
Hence, from the above,
We can conclude that the solutions for the given equation are: 4 and -10

Question 21.
x2 + 2x + 5 = 4
Answer:
The given equation is:
x² + 2x + 5 = 4
Subtract with 5 on both sides
So,
x²+ 2x + 5 – 5 = 4 – 5
x²+ 2x = -1
Compare the above equation with
ax² + bx = d
We know that,
To complete the square,
We have to know that value of c
So,
c = (\(\frac{b}{2}\))²
So,
The completed equation will be in the form of
ax² + bx + c = d
So,
x² + 2x + (\(\frac{2}{2}\))² = -1 + (\(\frac{2}{2}\))²
x² + 2x + 1 = -1 + 1
(x + 1)² = 0
√(x + 1)² = √0
x + 1 = 0
x = -1
Hence, from the above,
We can conclude that the solution for the given equation is: -1

Question 22.
2x2 – 4x = 10
Answer:
The given equation is:
2x² – 4x = 10
Divide by 2 into both sides
So,
x² – 2x = 5
Compare the above equation with
ax² + bx = d
So,
To complete the square,
We have to know that value of c
So,
c = (\(\frac{b}{2}\))²
So,
The completed equation will be in the form of
ax² + bx + c = d
So,
x² – 2x + 1 = 5 + 1
(x – 1)² = 6
√(x – 1)² = √6
x – 1 = ±√6
We know that,
√6 = 2.44
So,
x = 2.44 + 1 or x = -2.44 + 1
x = 3.44 or x = -1.44
Hence, from the above,
We can conclude that the solutions for the given equation are: 3.44 and -1.44

Determine whether the quadratic function has a maximum or minimum value. Then find the value.
Question 23.
y = -x2 + 6x – 1
Answer:
The given equation is:
y = -x²+ 6x – 1
Add with 1 on both sides
So,
y + 1 = -x² + 6x – 1 + 1
y + 1 = -(x² – 6x)
To complete the square,
We have to know the value of c
So,
c = (\(\frac{b}{2}\))²
So,
y + 1 – 9 = -(x² – 6x + 9)
y – 8 = -(x – 3)²
y = -(x – 3)² + 8
From the above vertex form,
We can observe that the value of a is negative
Since a is negative, the parabola will be closed down and the value of y becomes the maximum value
Hence, from the above,
We can conclude that the given function has a maximum value and the maximum value is: 8

Question 24.
f(x) = x2 + 4x + 11
Answer:
The given function is:
f (x) = x² + 4x + 11
Subtract with 11 on both sides
f (x) – 11 = x² + 4x + 11 – 11
f (x) – 11 = x² + 4x
Compare the above equation with
ax² + bx = d
To complete the square,
We have to know the value of c
So,
c = (\(\frac{b}{2}\))²
So,
f (x) – 11 + 4 = x² + 4x + 4
f (x) – 7 = (x + 2)²
f (x) = (x + 2)² + 7
From the above vertex form,
We can observe that the value of a is positive
Since a is positive, the parabola will be open up and the value of y becomes the minimum value
Hence, from the above,
We can conclude that the given function has a minimum value and the minimum value is: 7

Question 25.
y = 3x2 – 24x + 15
Answer:
The given equation is
y = 3x² – 24x + 15
Subtract with 15 on both sides
So,
y – 15 = 3x² – 24x + 15 – 15
y – 15 = 3x² – 24x
y – 15 = 3 (x² – 8x)
Compare the above equation with
ax² + bx = d
To complete the square,
We have to know that value of c,
So,
c = (\(\frac{b}{2}\))²
y – 15 + 144 = 3(x² – 8x + 48)
y + 129 = 3 (x – 12)²
y = 3 (x – 12)² – 129
From the above vertex form,
We can observe that the value of a is negative
Since a is negative, the parabola will be closed down and the value of y becomes the maximum value
Hence, from the above,
We can conclude that the given function has a maximum value and the maximum value is: -129

Question 26.
The width w of a credit card is 3 centimeters shorter than the length ℓ. The area is 46.75 square centimeters. Find the perimeter.
Answer:
It is given that the width w of a credit card is 3 centimeters shorter than the length ℓ. The area is 46.75 square centimeters.
So,
Let the length of a credit card be x cm
So,
The width of a credit card is: (x – 3) cm
So,
46.75 = x (x – 3)
x² – 3x = 46.75
x² – 3x – 46.75 = 0

So,
From the graph,
We can observe that
The solutions of x are: -6 and 8
We know that,
The length will not be negative,
So,
The length of the credit card is: 8 cm
The width of the credit card is: 8 – 3 = 5 cm
So,
The perimeter of the credit card = 2 (l + b)
= 2 (8 + 5) = 2 (13) = 26 cm
Hence, from the above,
We can conclude that the perimeter of the credit card is: 26 cm

9.5 Solving Quadratic Equations Using the Quadratic Formula (pp. 515–524)

Solve the equation using the Quadratic Formula. Round your solutions to the nearest tenth, if necessary.
Question 27.
x2+ 2x – 15 = 0
Answer;
The given equation is:
x² + 2x – 15 = 0
Compare the given equation with
ax²+ bx + c = 0
So,
a = 1, b = 2, and c = -15
So,
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{-2 ± \sqrt{2² + 4 (1) (15)}}{2 (1)}\)
x = \(\frac{-2 ± \sqrt{64}}{2}\)
x = \(\frac{-2 ± 8}{2}\)
x = 3 or x = -5
Hence, from the above,
We can conclude that the solutions for the given equation are: 3 and -5

Question 28.
2x2 – x + 8 = 16
Answer:
The given equation is:
2x² – x + 8 = 0
Compare the given equation with
ax²+ bx + c = 0
So,
a = 2, b = -1, and c = 8
So,
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{1 ± \sqrt{(-1)² – 4 (2) (8)}}{2 (2)}\)
x = \(\frac{1 ± \sqrt{-63}}{4}\)
We know that,
The square of a real number will not be a negative number
Hence, from the above,
We can conclude that there are no real solutions for the given equation

Question 29.
-5x2 + 10x = 5
Answer;
The given equation is:
-5x² + 10x = 5
So,
5x² – 10x + 5 = 0
Divide by 2 on both sides
So,
x² – 2x + 1 = 0
Compare the given equation with
ax²+ bx + c = 0
So,
a = 1, b = -2, and c = 1
So,
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{2 ± \sqrt{(-2)² – 4 (1) (1)}}{2 (1)}\)
x = \(\frac{2 ± \sqrt{0}}{2}\)
x = \(\frac{2}{2}\)
x = 1
Hence, from the above,
We can conclude that the solution for the given equation is: 1

Find the number of x-intercepts of the graph of the function.
Question 30.
y = -x2 + 6x – 9
Answer:
The given equation is:
y = -x² + 6x – 9
To find the x-interxept,
Put y = 0
So,
-x² + 6x – 9 = 0
x² – 6x + 9 = 0
x² – 3x – 3x + 9 = 0
x (x – 3) – 3 (x – 3) = 0
(x – 3) (x – 3) = 0
x – 3 = 0 or x – 3 = 0
x = 3 or x = 3
Hence, from the above,
We can conclude that the x-intercept for the given equation is: 3

Question 31.
y = 2x2 + 4x + 8
Answer:
The given equation is:
y = 2x² + 4x + 8
To find the x-intercept,
Put y = 0
So,
2x² + 4x + 8 = 0
Divide by 2 on both sides
x² + 2x + 4 = 0
Compare the given equation with
ax² + bx + c = 0
So,
a = 1, b = 2, and c = 4
So,
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{-2 ± \sqrt{2² – 4 (1) (4)}}{2 (1)}\)
x = \(\frac{-2 ± \sqrt{-12}}{2a}\)
We know that,
The square of a real number will not take any negative values
Hence, from the above,
We can conclude that there are no solutions for the given equation

Question 32.
y = – \(\frac{1}{2}\)x2 + 2x
Answer:
The given equation is:
y = –\(\frac{1}{2}\)x² + 2x
To find the x-intercept,
Put y = 0
So,
– \(\frac{1}{2}\)x2 + 2x = 0
– \(\frac{1}{2}\)x2 = – 2x
\(\frac{1}{2}\)x2 = 2x
x = 2 (2)
x = 4
Hence, from the above,
We can conclude that the solution for the given equation is: 4

9.6 Solving Nonlinear Systems of Equations (pp. 525–532)

Solve the system using any method.
Question 33.
y = x2 – 2x – 4
y = -5
Answer:
The given system of equations is:
y = x² – 2x – 4
y = 5
So,
x² – 2x – 4 = 5
x² – 2x = 5 + 4
x² – 2x = 9
So,
The representation of the given system of equations in the coordinate plane is:

Hence, from the above,
We can conclude that the solutions for the given system of equations are: -2 and 4

Question 34.
y = x2 – 9
y = 2x + 5
Answer:
The given system of equations are:
y = x² – 9
y = 2x + 5
So,
x² – 9 = 2x + 5
x² – 2x – 9 – 5 = 0
x² – 2x – 14 = 0

We know that,
√5 = 3.87
So,
x = -2.87 or x = 4.87
Hence, from the above,
We can conclude that the solutions for the given system of equations are: 4.87 and -2.87

Question 35.
y = 2(\(\frac{1}{2}\))x – 5
y = -x2 – x + 4
Answer:

Solving Quadratic Equations Chapter Test

Solve the equation using any method. Explain your choice of method.
Question 1.
x2 – 121 = 0
Answer:
The given equation is:
x² = 121 = 0
Add with 121 on both sides
So,
x² – 121 + 121 = 0 + 121
x² = 121
√x² = √121
x = ±11
Hence, from the above,
We can conclude that the solutions for the given equation are: 11 and -11

Question 2.
x2 – 6x = 10
Answer:
The given equation is:
x² – 6x = 10
So,
By using the method of completing the squares,
x² – 6x + 9 = 10 + 9
(x – 3)² = 19
√(x – 3)² = √19
x – 3 = ±√19
We know that,
√19 = 4.35
So,
x = 4.35 + 3 or x = -4.35 + 3
x = 7.35 or x = –1.35
Hence, from the above,
We can conclude that the solutions for the given equation are: 7.35 and -1.35

Question 3.
-2x2 + 3x + 7 = 0
Answer:
The given equation is:
-2x² + 3x + 7 = 0
Compare the given equation with
ax²+ bx + c = 0
So,

We know that,
√65 = 8.06
So,
x = -1.26 or x = 2.76
Hence, from the above,
We can conclud ethat the solutions for the given equation are: -1.26 and 2.76

Question 4.
x2 – 7x + 12 = 0
Answer:
The given equation is:
x² – 7x + 12 = 0
So,
x² – 4x – 3x + 12 = 0
x (x – 4) – 3 (x – 4) = 0
(x – 3) (x – 4) = 0
x – 3 = 0 or x – 4 = 0
x = 3 or x = 4
Hence, from the above,
We can conclude that the solutions for the given equation are: 3 and 4

Question 5.
5x2 + x – 4 = 0
Answer:
The given equation is:
5x² + x – 4 = 0
So,
5x² + 5x – 4x – 4 = 0
5x (x + 1) – 4 (x + 1) = 0
(5x – 4) (x + 1) = 0
5x – 4 = 0 or x + 1 = 0
5x = 4 or x = -1
x = \(\frac{4}{5}\) or x = -1
Hence, from the above,
We can conclude that the solutions for the given equation are: \(\frac{4}{5}\) and -1

Question 6.
(4x + 3)2 = 16
Answer:
The given equation is:
(4x + 3)² = 16
So,
√(4x + 3)² = √16
4x + 3 = 4 or 4x + 3 = -4
4x = 4 – 3 or 4x = -4 – 3
4x = 1 or 4x = -7
x = \(\frac{1}{4}\) or x = –\(\frac{7}{4}\)
Hence, from the above,
We can conclude that the solutions for the given equation are: \(\frac{1}{4}\) and –\(\frac{7}{4}\)

Question 7.
Describe how you can use the method of completing the square to determine whether the function f(x) = 2x2 + 4x – 6 can be represented by the graph shown.
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations ct 7
Answer:
The given equation is:
f (x) = 2x² + 4x – 6
To find the solutions,
We have to make f (x) = 0
So,
2x²+ 4x – 6 = 0
Add with 6 on both sides
2x²+ 4x – 6 + 6 = 0 + 6
2x² + 4x = 6
Now,
By using the method of completing the squares,
Step 1:
Divide with x² coefficient
Divide by 2 on both sides
So,
x²+ 2x = 3
Step 2:
Find the one half of the value of b
So,
\(\frac{b}{2}\) = \(\frac{2}{2}\)
= 1
Step 3:
Add the square of the result that we obtained in step 2 in the above equation
So,
Add 1 on both sides
So,
x²+ 2x + 1 = 3 + 1
(x + 1)² = 4
√(x + 1)² = √4
x + 1 = 2 or x + 1 = -2
x = 2 – 1 or x = -2 – 1
x = 1 or x = -3
Hence, from the above,
We can conclude that the solutions for the given equation are: 1 and -3

Question 8.
Write an expression involving radicals in which a conjugate can be used to simplify the expression.
Answer:
The expression that involves radicals in which a conjugate can be used to simplify the expression is:
\(\frac{8}{3 + \sqrt{6}}\)
The conjugate of the above expression is:
\(\frac{3 – \sqrt{6}}\)

Solve the system using any method.
Question 9.
y = x2 – 4x – 2
y = -4x + 2
Answer:
The given system of equations are:
y = x² – 4x – 2
y = -4x + 2
So,
x² – 4x – 2 = -4x + 2
x² – 4x – 2 + 4x – 2 = 0
x² – 4 = 0
x² = 4
√x² = √4
x = 2 or x = -2
Hence, from the above,
We can conclude that the solutions for the given set of equations are: 2 and -2

Question 10.
y = -5x2 + x – 1
y = -7
Answer:
The given system of equations are:
y = -5x²+ x – 1
y = -7
So,
-5x² + x – 1 = -7
-5x² + x – 1 + 7 = 0
-5x² +x + 6 = 0
-5x² + 6x – 5x + 6 = 0
-5x(x + 1) + 6 (x + 1) = 0
(-5x + 6) (x + 1) = 0
-5x + 6 = 0 or x + 1 = 0
-5x = -6 or x = -1
x = \(\frac{6}{5}\) or x = -1
Hence, from the above,
We can conclude that the solutions for the given set of equations are: \(\frac{6}{5}\) and -1

Question 11.
y = \(\frac{1}{2}\)(4)x + 1
y = x2 – 2x + 4
Answer:
The given system of equations are:
y = \(\frac{1}{2}\) (4)^x + 1
y = x² – 2x + 4
So,
The representation of the given system of equations in the coordinate plane is:

Hence, from the above,
We can conclude that the solution for the given set of equations is: (3, 1)

Question 12.
A skier leaves an 8-foot-tall ramp with an initial vertical velocity of 28 feet per second. The function h = -16t2 + 28t + 8 represents the height h (in feet) of the skier after t seconds. The skier has a perfect landing. How many points does the skier earn?
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations ct 12
Answer:

Hence, from the above,
We can conclude that the number of points the skier can earn is: 35.875 points

Question 13.
An amusement park ride lifts seated riders 265 feet above the ground. The riders are then dropped and experience free fall until the brakes are activated 105 feet above the ground. The function h = -16t2 + 265 represents the height h (in feet) of the riders t seconds after they are dropped. How long do the riders experience free fall? Round your solution to the nearest hundredth.
Answer:

Question 14.
Write an expression in the simplest form that represents the area of the painting shown.
Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations ct 14
Answer:
From the given figure,
We can observe that the painting is in the form of a rectangle
So,
The area of the painting = \(\sqrt{30 x^{7}}\) ×\(\frac{36}{\sqrt{3}}\)
= \(\sqrt{10 (3) x^{6} x}\) ×\(\frac{36}{\sqrt{3}}\)
= x³ \(\sqrt{10}\) × \(\sqrt{3}\) × \(\frac{36}{\sqrt{3}}\)
= 36x³ \(\sqrt{10}\)
Hence, from the above,
We can conclude that the area of the painting in the simplest form is: 36x³ \(\sqrt{10}\)

Question 15.
Explain how you can determine the number of times the graph of y = 5x2 – 10x + 5 intersects the x-axis without graphing or solving an equation.
Answer:
The given equation is:
y = 5x² – 10x + 5
We can determine the number of times the graph of the given equation intersects the x-axis by using the discriminant
We know that,
Discriminant (d) = b² – 4ac
So,
b² – 4ac > 0 ——> 2 intersection points
b² – 4ac = 0 ——-> 1 intersection point
b² – 4ac < 0 ——-> No intersection points
Now,
Compare the given equation with
ax² + bx + c = 0
So,
a = 5, b = -10, and c = 5
So,
b² – 4ac = (-10)² – 4 (5) (5)
= 100 – 100
= 0
Hence, from the above,
We can conclude that the given equation will have only 1 intersection point

Question 16.
Consider the quadratic equation ax2 + bx + c = 0 Find values of a, b, and c so that the graph of its related function has (a) two  x-intercepts, (b) one x-intercepts.and (c) no x-intercepts.
Answer:
The given equation is:
ax² + bx + c = 0
We know that,
If
b² – 4ac > 0 ——> 2 x-intercepts
b² – 4ac = 0 ——-> 1 x-intercept
b² – 4ac < 0 ——-> No x-intercepts
So,
a) For 2 x-intercepts:
b² – 4ac > 0
b² > 4ac
4ac < b²
c < \(\frac{b²}{4a}\)
b) For 1 x-intercept:
b² – 4ac = 0
b² = 4ac
4ac = b²
c = \(\frac{b²}{4a}\)
c) For no x-intercepts:
b² – 4ac < 0
b² < 4ac
4ac > b²
c > \(\frac{b²}{4a}\)

Question 17.
The numbers y of two types of bacteria after x hours are represented by the models below.
y = 3x2 + 8x + 20 Type A
y = 27x + 60 Type B
a. When are there 400 Type A bacteria?
b. When are the number of Type A and Type B bacteria the same?
c. When are there more Type A bacteria than Type B? When are there more Type B bacteria than Type A? Use a graph to support your answer.
Answer:

Solving Quadratic Equations Cumulative Assessment

Question 1.
The graphs of four quadratic functions are shown. Determine whether the discriminants of the equations formed by setting each function equal to zero are positive, negative, or zero.f(x) = 0, g(x) = 0, h(x) = 0, and j(x) = 0
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations ca 1
Answer:
We know that,
Discriminant (d) = b² – 4ac
If
d > 0, then the equation will have 2 solutions
If d = 0, then the equation will have 1 solution
If d < 0, then the equation will have no solutions
So,
From the given graph,
For f (x) = 0,
f (x) will not have any solutions since it did not touch the x-axis
For g (x) = 0,
g (x) will have 2 solutions since it touched the x-axis at 2 points
For h (x) = 0,
h (x) will have only 1 solution since it touched the x-axis at only 1 point
For j (x) = 0,
j (x) will have only 1 solution since it touched the x-axis at only 1 point

Question 2.
The function f(x) = a(1.08)x represents the total amount of money (in dollars) in Account A after x years. The function g(x) = 600(b)x represents the total amount of money (in dollars) in Account B after x years. Fill in values for a and b so that each statement is true.
a. When a = ____ and b = ____, Account B has a greater initial amount and increases at a faster rate than Account A.
b. When a = ____ and b = ____, Account B has a lesser initial amount than Account A but increases at a faster rate than Account A.
c. When a = ____ and b = ____, Account B and Account A have the same initial amount, and Account B increases at a slower rate than Account A.
Answer:

Question 3.
Your friend claims to be able to find the radius r of each figure, given the surface area S. Do you support your friend’s claim? Justify your answer.
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations ca 3
Answer:
The claim of your friend is that we can find the radius r of each figure given the Surface area S
But,
We know that,
For some Surface areas, the variable “Height (h)” will be present
So,
We can find the surface area given radius r only in some cases
Hence, from the above,
We can conclude that the claim of your friend is not correct

Question 4.
The tables represent the numbers of items sold at a concession stand on days with different average temperatures. Determine whether the data represented by each table show a positive, negative, or no correlation.
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations ca 4
Answer:

Question 5.
Which graph shows exponential growth?
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations ca 5
Answer:
We know that,
The graph of an “Exponential function” will gradually increase or decrease
So,
From the above graphs,
Graph A shows the gradual increase
Graph B shows the gradual decrease
Hence, from the above,
We can conclude that graph A and graph B show the exponential growth

Question 6.
Which statement best describes the solution(s) of the system of equations?
y = x2 + 2x – 8
y = 5x + 2
A. The graphs intersect at one point, (-2, -8). So, there is one solution.
B. The graphs intersect at two points, (-2, -8) and (5, 27). So, there are two solutions.
C. The graphs do not intersect. So, there is no solution.
D. The graph of y = x2 + 2x – 8 has two x-intercepts. So, there are two solutions.
Answer:
The given system of equations are:
y = x² + 2x – 8
y = 5x + 2
So,
x² + 2x – 8 = 5x + 2
x² + 2x – 5x – 8 – 2 = 0
x² – 3x – 10 = 0
x² – 5x + 2x – 10 = 0
x (x – 5) + 2 (x – 5) = 0
(x + 2) (x – 5) = 0
x + 2 = 0 or x – 5 = 0
x = -2 or x = 5
So,
There are 2 x-intercepts for the given equation i.e., -2 and 5
Hence, from the above,
We can conclude that option D best describes the solutions of the given system of equations

Question 7.
Which expressions are in the simplest form?
Big Ideas Math Algebra 1 Solutions Chapter 9 Solving Quadratic Equations ca 7
Answer:
Let the given expressions be named as a, b, c, d, e, f, g, h, i, and j respectively
Hence, from the above,
We can conclude that the expressions that are in the simplest form are: b, d, e, g, i, and j

Question 8.
The domain of the function shown is all integers in the interval -3 < x ≤ 3. Find all the ordered pairs that are solutions of the equation y = f(x).
f(x) = 4x – 5
Answer:
The given equation is:
f (x) = 4x – 5
It is given that the domain of the given function is:
-3 < x ≤ 3
So,
The domain of the given function = (-2, -1, 0, 1, 2, 3)
Now,
For x = -2, f (-2) = -13
For x = -1, f (-1) = -9
For x = 0, f (0) = -5
For x = 1, f (1) = -1
For x = 2, f (2) = 3
For x = 3, f (3) = 7
Hence, from the above,
We can conclude that the ordered pairs for the given function are:
(-2, 13), (-1, -9), (0, -5), (1, -1), (2, 3), and (3, 7)

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