# Eureka Math Algebra 1 Module 4 Lesson 6 Answer Key

## Engage NY Eureka Math Algebra 1 Module 4 Lesson 6 Answer Key

### Eureka Math Algebra 1 Module 4 Lesson 6 Example Answer Key

Example 1.
A physics teacher put a ball at the top of a ramp and let it roll down toward the floor. The class determined that the height of the ball could be represented by the equation h = – 16t2 + 4, where the height, h, is measured in feet from the ground and time, t, is measured in seconds.

a. What do you notice about the structure of the quadratic expression in this problem?
There is no linear term, just a square and constant.

b. In the equation, explain what the 4 represents.
The height when the time is 0 (i.e., the initial height of the top of the ramp is 4 feet).

c. Explain how you would use the equation to determine the time it takes the ball to reach the floor.
The ball reaches the ground when the height is zero, so set the expression equal to zero.

d. Now consider the two solutions for t. Which one is reasonable? Does the final answer make sense based on this context? Explain.
Only t = + $$\frac{1}{2}$$ makes sense since time cannot be negative in this context. This means that it took $$\frac{1}{2}$$ sec. for the ball to travel to the end of the ramp. That would make sense if the ramp was pretty short.

Example 2.
Lord Byron is designing a set of square garden plots so some peasant families in his kingdom can grow vegetables. The minimum size for a plot recommended for vegetable gardening is at least 2 m on each side. Lord Byron has enough space around the castle to make bigger plots. He decides that each side should be the minimum (2 m) plus an additional x m.
a. What expression can represent the area of one individual garden based on the undecided additional length x?
(x + 2)2

b. There are 12 families in the kingdom who are interested in growing vegetables in the gardens. What equation can represent the total area, A, of the 12 gardens?
A = 12 = (x + 2) = 2

c. If the total area available for the gardens is 300 sq m, what are the dimensions of each garden?
12(x + 2)2 = 300
(x + 2)2 = 25 (Consider and discuss why we divide both sides of the equation by 12 BEFORE we take the square root.)
(x + 2) = 5 or – 5. The side length for the garden is 5 m. (Note: Make sure to emphasize the rejection of the – 5 in this context (the length of the side is given as x + 2, which cannot be negative) but also to point out that not ALL negative solutions are rejected for ALL problems in a context.)

d. Find both values for x that make the equation in part (c) true (the solution set). What value of x does Lord Byron need to add to the 2 m? (x + 2) = 5 or – 5, so x = 3 or – 7.
He needs to add 3 m to the minimum measurement of 2 m.

### Eureka Math Algebra 1 Module 4 Lesson 6 Exercise Answer Key

Exercises
Solve each equation. Some of them may have radicals in their solutions.
Exercise 1.
3x2 – 9 = 0
3x2 = 9 → x2 = 3 → x = ±$$\sqrt{3}$$

Exercise 2.
(x – 3)2 = 1
(x – 3) = ±1 → x = 3±1 → x = 2 or 4

Exercise 3.
4(x – 3)2 = 1
(x – 3)2 = $$\frac{1}{4}$$ → (x – 3) = ± $$\frac{1}{2}$$ → x = 3±$$\frac{1}{2}$$ → x = $$\frac{7}{2}$$ or $$\frac{5}{2}$$

Exercise 4.
2(x – 3)2 = 12
(x – 3)2 = 6 → (x – 3) = ±$$\sqrt{6}$$ → x = 3±$$\sqrt{6}$$ (As estimated decimals: 5.45 or 0.55)

Exercise 5.
Analyze the solutions for Exercises 2–4. Notice how the questions all had (x – 3)2 as a factor, but each solution was different (radical, mixed number, whole number). Explain how the structure of each expression affected each problem – solution pair.
Question 2: In the equation, (x – 3)2 equals a perfect square (1). When the square root is taken, we get x – 3 = ±1, which yields whole – number solutions.
Question 3: After both sides are divided by 4, 4(x – 3)2 = 1 becomes (x – 3)2 = $$\frac{1}{4}$$, which is a fraction that has a perfect square for the numerator and denominator. Therefore, when the square root is taken, we get x – 3 = ±$$\frac{1}{2}$$, which yields fraction solutions.
Question 4: In this equation, (x – 3)2 does not equal a perfect square or a fraction whose denominator and numerator are perfect squares after both sides are divided by 2. Instead, x – 3 equals an irrational number after we take the square root of both sides.

Exercise 6.
Peter is a painter, and he wonders if he would have time to catch a paint bucket dropped from his ladder before it hits the ground. He drops a bucket from the top of his 9 – foot ladder. The height, h, of the bucket during its fall can be represented by the equation, h = – 16t2 + 9, where the height is measured in feet from the ground, and the time since the bucket was dropped, t, is measured in seconds. After how many seconds does the bucket hit the ground? Do you think he could catch the bucket before it hits the ground?
– 16t2 + 9 = 0 → – 16t2 = – 9 → t2 = – $$\frac{9}{ – 16}$$ → t2 = $$\frac{9}{16}$$ → t = ±$$\frac{3}{4}$$
The bucket will hit the ground after $$\frac{3}{4}$$ seconds. I do not think he could catch the bucket before it hits the ground. It would be impossible for him to descend the 9 – foot ladder and catch the bucket in $$\frac{3}{4}$$ seconds.

### Eureka Math Algebra 1 Module 4 Lesson 6 Problem Set Answer Key

Question 1.
Factor completely: 15x2 – 40x – 15.
GCF is 5: 5(3x2 – 8x – 3) = 5(3x + 1)(x – 3).

Solve each equation.
Question 2.
4x2 = 9
x2 = $$\frac{9}{4}$$ → x = ±$$\frac{3}{2}$$

Question 3.
3y2 – 8 = 13
3y2 = 21 → y2 = 7 → y = ±$$\sqrt{7}$$

Question 4.
(d + 4)2 = 5
d + 4 = ±$$\sqrt{5}$$ → d = – 4±$$\sqrt{5}$$

Question 5.
4(g – 1)2 + 6 = 13
4(g – 1)2 = 7 → (g – 1)2 = $$\frac{7}{4}$$ → g – 1 = ±$$\frac{\sqrt{7}}{2}$$ → g = 1 ±$$\frac{\sqrt{7}}{2}$$

Question 6.
12 = – 2(5 – k)2 + 20
– 8 = – 2(5 – k)2 → 4 = (5 – k)2 → (5 – k) = ± 2 → – k = – 5 ± 2 = – 3 or – 7, so k = 3 or 7

Question 7.
Mischief is a toy poodle that competes with her trainer in the agility course. Within the course, Mischief must leap through a hoop. Mischief’s jump can be modeled by the equation h = – 16t2 + 12t, where h is the height of the leap in feet and t is the time since the leap, in seconds. At what values of t does Mischief start and end the jump?
To find the start and end of the jump, we need to find where height,h, is zero and solve the resulting equation.
– 16t2 + 12t = 0
– 4t(4t – 3) = 0
t = 0 or $$\frac{3}{4}$$ seconds
The leap starts at 0 seconds and ends at $$\frac{3}{4}$$ seconds.
(Students may decide to factor the GCF, – 16t, for the factoring step and obtain – 16t(t – $$\frac{3}{4}$$) = 0. They should still arrive at the same conclusion.)

### Eureka Math Algebra 1 Module 4 Lesson 6 Exit Ticket Answer Key

Question 1.
Solve the equations.
a. 4a2 = 16
a2 = 4
a = 2 or – 2

b. 5b2 – 25 = 0
5b2 = 25
b2 = 5
b = ±$$\sqrt{5}$$

c. 8 – c2 = 5
– c2 = – 3
c2 = 3
c = ±$$\sqrt{3}$$

Question 2.
Solve the equations.
a. (x – 2)2 = 9
(x – 2) = ± 3
x = 2±3 = – 1 or 5

b. 3(x – 2)2 = 9
x – 2 = ±$$\sqrt{3}$$
x = 2 ±$$\sqrt{3}$$
(x + 1)2 = $$\frac{6}{24}$$ = $$\frac{1}{4}$$
x + 1 = ±$$\sqrt{\frac{1}{4}}$$ = ±$$\frac{1}{2}$$
x = – 1 ±$$\frac{1}{2}$$
x = – $$\frac{1}{2}$$ or – $$\frac{3}{2}$$