## 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: x^{2} + 2x = 8.

Answer:

To solve by completing the square, factor out the leading coefficient, complete the square, and balance the equality.

(x^{2} + 2x + ) = 8 +

(x^{2} + 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: 7p^{2} – 12p + 4 = 0.

Answer:

To solve by completing the square, first gather the variable terms to one side of the equation and factor out the leading coefficient.

7p^{2} – 12p = – 4

7(p^{2} – \(\frac{12}{7}\) 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 ax^{2} + bx + c = 0.

Answer:

Exercises 1–4

Use the quadratic formula to solve each equation.

Exercise 1.

x^{2} – 2x = 12→a = 1, b = – 2, c = – 12 [Watch the negatives.]

Answer:

Exercise 2.

\(\sqrt{\frac{1}{9}}\) r^{2} – 6r = 2→a = \(\frac{1}{2}\), b = – 6, c = – 2 [Did you remember the negative?]

Answer:

Exercise 3.

2p^{2} + 8p = 7→a = 2, b = 8, c = – 7

Answer:

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.

2y^{2} + 3y – 5 = 4→a = 2, b = 3, c = – 9

Answer:

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 ______________

2x^{2} + 5x – 3 = 0

Method ______________

x^{2} + 3x – 5 = 0

Method ______________

\(\frac{1}{2}\) x^{2} – x – 4 = 0

Answer:

(Students may choose any method to solve. However, the first and third are factorable, but the middle expression is not.)

Method by factoring

2x^{2} + 5x – 3 = 0

(2x – 1)(x + 3) = 0

x = \(\frac{1}{2}\) or – 3

Method by quadratic formula

x^{2} + 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

\(\frac{1}{2}\) x^{2} – x – 4 = 0

\(\frac{1}{2}\) (x^{2} – 2x ) = 4

\(\frac{1}{2}\) (x^{2} – 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: z^{2} – 3z – 8 = 0.

Answer:

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: 2q^{2} – 8 = 3q.

Answer:

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: \(\frac{1}{3}\) m^{2} + 2m + 8 = 5.

Answer:

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.

\(\frac{3}{2}\) R^{2} – 2R = 2

Answer:

Let’s start each method by multiplying both sides of the equation by 2 to eliminate the fraction.

3R^{2} – 4R = 4