Engage NY Eureka Math Algebra 1 Module 4 Lesson 13 Answer Key
Eureka Math Algebra 1 Module 4 Lesson 13 Example Answer Key
Example 1
Solve for x.
12 = x2 + 6x
Answer:
x2 + 6x + 9 = 12 + 9 Add 9 to complete the square: [\(\frac{1}{2}\) (6)]2.
(x + 3)2 = 21 Factor the perfect square.
x + 3 = ±\(\sqrt{21}\) Take the square root of both sides. Remind students NOT to forget the ±.
x = – 3 ± \(\sqrt{21}\) Add – 3 to both sides to solve for x.
x = – 3 + \(\sqrt{21}\) or x = – 3 – \(\sqrt{21}\)
→ Remember to put the – 3 to the left of the ± square root.
→ Is there a simpler way to write this answer?
→ No, since the value under the radical is not a perfect square, it cannot be written in a simpler form or combined with the – 3 by addition or subtraction. The number cannot be expressed exactly or in any simpler form.
→ The solutions can also be estimated using decimals, approximately 1.58 or – 7.58.
Example 2.
Solve for x.
4x2 – 40x + 94 = 0
Answer:
4x2 – 40x + 94 = 0 or 4x2 – 40x = – 94
Gathering everything on the left:
4(x2 – 10x + ____) + 94 = 0
4(x2 – 10x + 25) + 94 – 100 = 0
4(x – 5)2 – 6 = 0
4(x – 5)2 = 6
(x – 5)2 = \(\frac{6}{4}\)
x – 5 = ±\(\sqrt{\frac{6}{4}}\)
x = 5±\(\frac{\sqrt{6}}{2}\)
x = 5 + \(\frac{\sqrt{6}}{2}\) or 5 – \(\frac{\sqrt{6}}{2}\)
Gathering variable terms on the left and the constant on the right:
4(x2 – 10x + ____) = – 94
4(x2 – 10x + 25) = – 94 + 100
4(x – 5)2 = 6
(x – 5)2 = \(\frac{6}{4}\)
x – 5 = ±\(\sqrt{\frac{6}{4}}\)
x = 5±\(\frac{\sqrt{6}}{2}\)
x = 5 + \(\frac{\sqrt{6}}{2}\) or 5 – \(\frac{\sqrt{6}}{2}\)
Eureka Math Algebra 1 Module 4 Lesson 13 Exercise Answer Key
Opening Exercise
a. Solve the equation for b: 2b2 – 9b = 3b2 – 4b – 14.
Answer:
To solve by factoring, gather all terms to one side of the equation and combine like terms so that the remaining expression is equal to zero:
b2 + 5b – 14 = 0, then factor: (b + 7)(b – 2) = 0, and solve: b = – 7 or 2.
b. Rewrite the expression by completing the square: \(\frac{1}{2}\) b2 – 4b + 13.
Answer:
Factor \(\frac{1}{2}\) from the first two terms: \(\frac{1}{2}\)(b2 – 8b ) + 13, and then complete the square by adding + 16 inside the parentheses (which is really + 8 since there is \(\frac{1}{2}\) outside the parentheses). Now, to compensate for the + 8, we need to add – 8 outside the parentheses and combine it with the constant term:
\(\frac{1}{2}\)(b2 – 8b + 16) + 13 – 8 = \(\frac{1}{2}\) (b – 4)2 + 5 .
Exercises
Solve each equation by completing the square.
Exercise 1.
x2 – 2x = 12
Answer:
x2 – 2x + 1 = 12 + 1
(x – 1)2 = 13
x = 1±\(\sqrt{13}\)
Exercise 2.
\(\frac{1}{2}\) r2 – 6r = 2
Answer:
\(\frac{1}{2}\)(r2 – 12r + 36) = 2 + 18 (Be careful with factoring out the rational leading coefficient.)
\(\frac{1}{2}\) (r – 6)2 = 20
(r – 6)2 = 40
r – 6 = ±\(\sqrt{40}\) (The last step should be optional at this point.)
r = 6±\(\sqrt{40}\) = 6±2\(\sqrt{10}\)
Exercise 3.
2p2 + 8p = 7
Answer:
2(p2 + 4p + 4) = 7 + 8
2(p + 2)2 = 15
(p + 2)2 = \(\frac{15}{2}\)
(p + 2) = ±\(\sqrt{\frac{15}{2}}\)
p = – 2±\(\sqrt{\frac{15}{2}}\); – 2 + \(\sqrt{\frac{15}{2}}\) or – 2 – \(\sqrt{\frac{15}{2}}\)
Exercise 4.
2y2 + 3y – 5 = 4
Answer:
2y2 + 3y = 4 + 5
2[y2 + (\(\frac{3}{2}\))y + \(\frac{9}{16}\)] = 9 + \(\frac{9}{8}\)
2(y + \(\frac{3}{4}\))2 = \(\frac{81}{8}\)
(y + \(\frac{3}{4}\))2 = \(\frac{81}{16}\)
(y + \(\frac{3}{4}\)) = ±\(\sqrt{\frac{81}{16}}\)
y = – \(\frac{3}{4}\)±\(\frac{9}{4}\)
y = \(\frac{3}{2}\) or – 3
(Notice the square in the numerator. It is best to leave the fraction as it is in this step since we know we will eventually be taking the square root.)
Eureka Math Algebra 1 Module 4 Lesson 13 Problem Set Answer Key
Solve each equation by completing the square.
Question 1.
p2 – 3p = 8
Answer:
p2 – 3p + \(\frac{9}{4}\) = 8 + \(\frac{9}{4}\)
(p – \(\frac{3}{2}\))2 = \(\frac{41}{4}\)
(p – \(\frac{3}{2}\)) = ±\(\sqrt{\frac{41}{4}}\))
p = \(\frac{3}{2}\) ± \(\frac{\sqrt{41}}{2}\)
Question 2.
2q2 + 8q = 3
Answer:
2(q2 + 4q + 4) = 3 + 8
2(q + 2)2 = 11
(q + 2)2 = \(\frac{11}{2}\)
(q + 2) = ±\(\sqrt{\frac{11}{2}}\)
q = – 2±\(\sqrt{\frac{11}{2}}\)
Question 3.
\(\frac{1}{3}\) m2 + 2m + 8 = 5
Answer:
\(\frac{1}{3}\) (m2 + 6m) + 8 – 8 = 5 – 8
\(\frac{1}{3}\) (m2 + 6m + 9) = – 3 + 3
\(\frac{1}{3}\) (m + 3)2 = 0
(m + 3)2 = 0
m = – 3
Question 4.
– 4x2 = 24x + 11
Answer:
– 4x2 – 24x = 11
– 4(x2 + 6x + 9) = 11 – 36
Gather variable terms.
Factor out – 4; complete the square and balance the equality.
Factor the perfect square.
Divide both sides by – 4.
– 4(x + 3)2 = – 25
(x + 3)2 = + \(\frac{25}{4}\)
x + 3 = ± \(\frac{5}{2}\)
x = – 3±\(\frac{5}{2}\) = – \(\frac{1}{2}\) or – 5 \(\frac{1}{2}\)
Eureka Math Algebra 1 Module 4 Lesson 13 Exit Ticket Answer Key
Question 1.
Solve the following quadratic equation both by factoring and by completing the square: \(\frac{1}{4}\) x2 – x = 3.
Answer:
Factoring—you can eliminate the fraction by multiplying both sides by 4 to obtain an equivalent expression.
x2 – 4x = 12
x2 – 4x – 12 = 0
(x – 6)(x + 2) = 0
x = 6 or – 2
Completing the Square―you can start by either multiplying both sides by 4 or by factoring out \(\frac{1}{4}\) as the GCF.
\(\frac{1}{4}\)(x2 – 4x + ____) = 3
\(\frac{1}{4}\)(x2 – 4x + 4) = 3 + 1
x2 – 4x + 4 = 16
(x – 2)2 = 16
x = 2 ± 4
x = 6 or – 2
Question 2.
Which method do you prefer to solve this equation? Justify your answer using algebraic reasoning.
Answer:
Students may have a personal preference about which way is easier; therefore, either answer is correct, but it should have mathematical reasoning to support it. They may prefer factoring since they are more familiar with it or because having a fraction as the leading coefficient can make completing the square trickier. Notice that in the factoring response, the fraction was eliminated in the first step. Remind students that this is often an option to make an equation look friendlier.