# Eureka Math Algebra 1 Module 4 Lesson 14 Answer Key

## Engage NY Eureka Math Algebra 1 Module 4 Lesson 14 Answer Key

### Eureka Math Algebra 1 Module 4 Lesson 14 Exercise Answer Key

Opening Exercise
a. Solve for x by completing the square: x2 + 2x = 8.
To solve by completing the square, factor out the leading coefficient, complete the square, and balance the equality.
(x2 + 2x + ) = 8 +
(x2 + 2x + 1) = 8 + 1
(x + 1)2 = 9
(x + 1) = ±$$\sqrt{9}$$
x = – 1±$$\sqrt{9}$$
x = – 1 + 3 or – 1 – 3
x = 2 or – 4
Note: This equation can also be solved by factoring.

b. Solve for p by completing the square: 7p2 – 12p + 4 = 0.
To solve by completing the square, first gather the variable terms to one side of the equation and factor out the leading coefficient.
7p2 – 12p = – 4
7(p2 – $$\frac{12}{7}$$ p + ($$\frac{6}{7}$$)2 ) = – 4 + $$\frac{7(36)}{49}$$
7(p–$$\frac{6}{7}$$)2 = – 4 + 3$$\frac{6}{7}$$
7(p – $$\frac{6}{7}$$ )2 = $$\frac{8}{7}$$
(p – $$\frac{6}{7}$$ )2 = $$\frac{8}{49}$$
(p – $$\frac{6}{7}$$) = ±$$\frac{\sqrt{8}}{7}$$
p = $$\frac{6}{7}$$ ± $$\frac{\sqrt{8}}{7}$$ OR approximately 1.26 or 0.45

Discussion
Solve ax2 + bx + c = 0. Exercises 1–4
Use the quadratic formula to solve each equation.
Exercise 1.
x2 – 2x = 12→a = 1, b = – 2, c = – 12 [Watch the negatives.] Exercise 2.
$$\sqrt{\frac{1}{9}}$$ r2 – 6r = 2→a = $$\frac{1}{2}$$, b = – 6, c = – 2 [Did you remember the negative?] Exercise 3.
2p2 + 8p = 7→a = 2, b = 8, c = – 7 Note: In the Lesson 13 problem, the radical in the final answer was $$\sqrt{\frac{15}{2}}$$, which is equivalent to $$\sqrt{\frac{30}{2}}$$.

Exercise 4.
2y2 + 3y – 5 = 4→a = 2, b = 3, c = – 9 Exercise 5.
Solve these quadratic equations, using a different method for each: solve by factoring, solve by completing the square, and solve using the quadratic formula. Before starting, indicate which method you will use for each.

Method ______________
2x2 + 5x – 3 = 0

Method ______________
x2 + 3x – 5 = 0

Method ______________
$$\frac{1}{2}$$ x2 – x – 4 = 0
(Students may choose any method to solve. However, the first and third are factorable, but the middle expression is not.)
Method by factoring
2x2 + 5x – 3 = 0
(2x – 1)(x + 3) = 0
x = $$\frac{1}{2}$$ or – 3

x2 + 3x – 5 = 0
x = $$\frac{ – 3 \pm \sqrt{3^{2} – 4(1)( – 5)}}{2(1)}$$
x = $$\frac{ – 3 \pm \sqrt{29}}{2}$$
x = $$\frac{ – 3 + \sqrt{29}}{2}$$ or $$\frac{ – 3 – \sqrt{29}}{2}$$
x≈1.19 or – 4.19

Method by completing the square
$$\frac{1}{2}$$ x2 – x – 4 = 0
$$\frac{1}{2}$$ (x2 – 2x ) = 4
$$\frac{1}{2}$$ (x2 – 2x + 1) = 4 + $$\frac{1}{2}$$
$$\frac{1}{2}$$ (x – 1)2 = 4.5
(x – 1)2 = 9
(x – 1) = ±3
x = 1±3
x = 4 or – 2

### Eureka Math Algebra 1 Module 4 Lesson 14 Problem Set Answer Key

Use the quadratic formula to solve each equation.
Question 1.
Solve for z: z2 – 3z – 8 = 0.
a = 1, b = – 3, c = – 8
z = $$\frac{ – ( – 3) \pm \sqrt{( – 3)^{2} – 4(1)( – 8)}}{2(1)}$$ = $$\frac{3 \pm \sqrt{41}}{2}$$

Question 2.
Solve for q: 2q2 – 8 = 3q.
a = 2, b = – 3, c = – 8
q = $$\frac{ – ( – 3) \pm \sqrt{( – 3)^{2} – 4(2)( – 8)}}{2(2)}$$ = $$\frac{3 \pm \sqrt{73}}{4}$$

Question 3.
Solve for m: $$\frac{1}{3}$$ m2 + 2m + 8 = 5.
a = $$\frac{1}{3}$$, b = 2, c = 3
m = $$\frac{ – 2 \pm \sqrt{2^{2} – 4\left(\frac{1}{3}\right)(3)}}{2\left(\frac{1}{3}\right)}$$ = $$\frac{ – 2 \pm \sqrt{0}}{\frac{2}{3}}$$ = – 3

### Eureka Math Algebra 1 Module 4 Lesson 14 Exit Ticket Answer Key

Question 1.
Solve for R using any method. Show your work.
$$\frac{3}{2}$$ R2 – 2R = 2 