Engage NY Eureka Math Algebra 2 Module 1 Lesson 9 Answer Key
Eureka Math Algebra 2 Module 1 Lesson 9 Example Answer Key
Example 1.
Express √50 – √18 + √8 in simplest radical form and combine like terms.
Answer:
√50 = \(\sqrt{25 \cdot 2}\) = √25 ∙ √2 = 5√2
√18 = \(\sqrt{9 \cdot 2}\) = √9 ∙ √2 = 3√2
√8 = \(\sqrt{4 \cdot 2}\) = √4 ∙ √2 = 2√2
Therefore, √50 – √18 + √8 = 5√2 – 3√2 + 2√2 = 4√2.
Example 2.
Multiply and combine like terms. Then explain what you notice about the two different results.
(√3 + √2) (√3 + √2)
(√3 + √2) (√3 – √2)
Answer:
Solution (with teacher comments and a question):
The first product is √3 ∙ √3 + 2(√3 ∙ √2) + √2 ∙ √2 = 5 + 2√6.
The second product is √3 ∙ √3 – (√3 ∙ √2) + (√3 ∙ √2 ) – √2 ∙ √2 = 3 – 2 = 1.
The first product is an irrational number; the second is an integer.
The second product has the nice feature that the radicals have been eliminated. In that case, the two factors are given a special name: two binomials of the form √a + √b and √a – √b are called conjugate radicals:
√a + √b is the conjugate of √a – √b and
√a – √b is the conjugate of √a + √b.
More generally, for any expression in two terms, at least one of which contains a radical, its conjugate is an expression consisting of the same two terms but with the opposite sign separating the terms. For example, the conjugate of 2 – √3 is 2 + √3, and the conjugate of \(\sqrt[3]{5}+\sqrt{3}\) is \(\sqrt[3]{5}-\sqrt{3}\).
→ What polynomial identity is suggested by the product of two conjugates?
Students should answer that it looks like the difference of two squares.
The product of two conjugates has the form of the difference of squares:
(x + y) (x – y ) = x2 – y2.
The following exercise focuses on the use of conjugates.
Example 3.
Write \(\frac{\sqrt{3}}{5-2 \sqrt{3}}\) in simplest radical form.
Answer:
Eureka Math Algebra 2 Module 1 Lesson 9 Opening Exercise Answer Key
Which of these statements are true for all a, b > 0? Explain your conjecture.
i. 2(a + b) = 2a + 2b
Answer:
ii. \(\frac{a+b}{2}\) = \(\frac{a}{2}+\frac{b}{2}\)
Answer:
iii. \(\sqrt{a+b}\) = \(\sqrt{a}+\sqrt{b}\)
Answer:
Eureka Math Algebra 2 Module 1 Lesson 9 Exercise Answer Key
Exercise 1.
\(\)
Answer:
2 – 3√5
Exercise 2.
√2 (√3 – √2)
Answer:
√6 – 2
Exercise 3.
\(\sqrt{\frac{3}{8}}\)
Answer:
\(\frac{\sqrt{3}}{\sqrt{8}}=\frac{\sqrt{6}}{\sqrt{16}}=\frac{\sqrt{6}}{4}\)
Exercise 4.
\(\sqrt[3]{\frac{5}{32}}\)
Answer:
\(\frac{\sqrt[3]{5}}{\sqrt[3]{32}}=\frac{\sqrt[3]{10}}{\sqrt[3]{64}}=\frac{\sqrt[3]{10}}{4}\)
Exercise 5.
\(\sqrt[3]{16 x^{5}}\)
Answer:
\(\sqrt[3]{8 x^{3}} \cdot \sqrt[3]{2 x^{2}}=2 x \sqrt[3]{2 x^{2}}\)
Exercise 6.
Find the product of the conjugate radicals.
(√5 + √3) (√5 – √3)
Answer:
5 – 3 = 2
(7 + √2) (7 – √2)
Answer:
49 – 2 = 47
(√5 + 2) (√5 – 2)
Answer:
5 – 4 = 1
Eureka Math Algebra 2 Module 1 Lesson 9 Problem Set Answer Key
Question 1.
Express each of the following as a rational number or in simplest radical form. Assume that the symbols a, b, and x represent positive numbers.
a. √36
Answer:
6
b. √72
Answer:
6√2
c. √18
Answer:
3√2
d. √9x3
Answer:
3x√x
e. √27x2
Answer:
3x√3
f. \(\sqrt[3]{16}\)
Answer:
\(2 \sqrt[3]{2}\)
g. \(\sqrt[3]{24 a}\)
Answer:
\(2 \sqrt[3]{3 a}\)
h. \(\sqrt{9 a^{2}+9 b^{2}}\)
Answer:
\(3 \sqrt{a^{2}+b^{2}}\)
Question 2.
Express each of the following in simplest radical form, combining terms where possible.
a. √25 + √45 − √20
Answer:
5 + √5
b. \(3 \sqrt{3}-\sqrt{\frac{3}{4}}+\sqrt{\frac{1}{3}}\)
Answer:
\(\frac{17 \sqrt{3}}{6}\)
c. \(\sqrt[3]{54}-\sqrt[3]{8}+7 \sqrt[3]{\frac{1}{4}}\)
Answer:
\(\frac{13 \sqrt[3]{2}}{2}-2\)
d. \(\sqrt[3]{\frac{5}{8}}+\sqrt[3]{40}-\sqrt[3]{\frac{8}{9}}\)
Answer:
\(\frac{5 \sqrt[3]{5}}{2}-\frac{2 \sqrt[3]{3}}{3}\)
Question 3.
Evaluate \(\sqrt{x^{2}-y^{2}}\) when x = 33 and y = 15.
Answer:
12√6
Question 4.
Evaluate \(\sqrt{x^{2}+y^{2}}\) when x = 20 and y = 10.
Answer:
10√5
Question 5.
Express each of the following as a rational expression or in simplest radical form. Assume that the symbols x and y represent positive numbers.
a. √3(√7 − √3)
Answer:
√21 − 3
b. (3 + √2)2
Answer:
11 + 6√2
c. (2 + √3) (2 − √3)
Answer:
1
d. (2 + 2√5) (2 − 2√5)
Answer:
– 16
e. (√7 − 3) (√7 + 3)
Answer:
– 2
f. (3√2 + √7) (3√2 − √7)
Answer:
11
g. (x − √3) (x + √3)
Answer:
x2 – 3
h. (2x√2 + y) (2x√2 – y)
Answer:
8x2 – y2
Question 6.
Simplify each of the following quotients as far as possible.
a. (√21 − √3) ÷ √3
Answer:
√7 − 1
b. (√5 + 4) ÷ (√5 + 1)
Answer:
\(\frac{1}{4}\) (1 + 3√5)
c. (3 − √2) ÷ (3√2 − 5)
Answer:
– \(\frac{1}{7}\) (9 + 4√2)
d. (2√5 − √3) ÷ (3√5 − 4√2)
Answer:
\(\frac{1}{13}\) (30 − 3√15 + 8√10 − 4√6)
Question 7.
If x = 2 + √3, show that x + \(\frac{1}{x}\) has a rational value.
Answer:
x + \(\frac{1}{x}\) = 4
Question 8.
Evaluate 5x2 – 10x when the value of x is \(\frac{2-\sqrt{5}}{2}\).
Answer:
\(\frac{5}{4}\)
Question 9.
Write the factors of a4 − b4. Express (√3 + √2)4 – (√3 − √2)4 in a simpler form.
Answer:
Factors: (a2 + b2) (a + b) (a – b)
Simplified form: 40√6
Question 10.
The converse of the Pythagorean theorem is also a theorem: If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle.
Use the converse of the Pythagorean theorem to show that for A, B, C > 0, if A + B = c, then √A + √B > √C, so that √A + √B > \(\sqrt{A+B}\).
Answer:
Solution 1: Since A, B, C > 0 we can interpret these quantities as the areas of three squares whose sides have lengths √A, √B, and √c. Because A + B = c, then by the converse of the Pythagorean theorem, √A √B, and √C, are the lengths of the legs and hypotenuse of a right triangle. In a triangle, the sum of any two sides is greater than the third side. Therefore, √A + √B > √C, so √A + √B > \(\sqrt{A+B}\).
Solution 2: Since A, B, C > 0, we can interpret these quantities as the areas of three squares whose sides have lengths √A, √B, and √C. Because A + B = C, then by the converse of the Pythagorean theorem, √A = a, √B = b, and c = √C are the lengths of the legs and hypotenuse of a right triangle, so a, b, c > 0 Therefore, 2ab > 0. Adding equal positive quantities to each side of that inequality, we get a2 + b2 + 2ab > c2, which we can rewrite as (a + b)2 > c2. Taking the positive square root of each side, we get a + b > c, or equivalently, √A + √B > √C.
We then have √A + √B > \(\sqrt{A+B}\).
Eureka Math Algebra 2 Module 1 Lesson 9 Exit Ticket Answer Key
Question 1.
Rewrite each of the following radicals as a rational number or in simplest radical form.
a. √49
Answer:
7
b. \(\sqrt[3]{40}\)
Answer:
\(2 \sqrt[3]{5}\)
c. √242
Answer:
11√2
Question 2.
Find the conjugate of each of the following radical expressions.
a. √5 + √11
Answer:
√5 − √11
b. 9 – √11
Answer:
9 + √11
c. \(\sqrt[3]{3}\) + 1.5
Answer:
\(\sqrt[3]{3}\) – 1.5
Question 3.
Rewrite each of the following expressions as a rational number or in simplest radical form.
a. √3 (√3 − 1)
Answer:
3 − √3
b. (5 + √3)2
Answer:
28 + 10√3
c. (10 + √11) (10 − √11)
Answer:
89