## 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
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
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.
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.
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
x2 – 2x + 1 = 12 + 1
(x – 1)2 = 13
x = 1±$$\sqrt{13}$$

Exercise 2.
$$\frac{1}{2}$$ r2 – 6r = 2
$$\frac{1}{2}$$(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
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
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
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
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
$$\frac{1}{3}$$ (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
– 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.
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.