## Engage NY Eureka Math Algebra 1 Module 1 Lesson 17 Answer Key

### Eureka Math Algebra 1 Module 1 Lesson 17 Example Answer Key

Examples 1–2
Work the two examples as a class.

Example 1.
Solve 2x2-10x=0, for x.
{0,5}

Example 2.
Solve x(x-3)+5(x-3)= 0, for x.
{-5,3}

Example 3.
Consider the equation (x-2)(2x-3)=(x-2)(x+5). Lulu chooses to multiply through by $$\frac{1}{x-2}$$ and gets the
answer x=8. But Poindexter points out that x=2 is also an answer, which Lulu missed.
a. What’s the problem with Lulu’s approach?
You cannot multiply by $$\frac{1}{x-2}$$ because x-2 could equal 0, which means that you would be dividing by 0.

b. Use factoring to solve the original equation for x.
(x-2)(2x-3)-(x-2)(x+5)=0
(x-2)(2x-3-(x+5))=0
(x-2)(x-8)=0
x-2=0 or x-8=0
x=2 or x=8

Work through the responses as a class.
→ Emphasize the idea that multiplying by $$\frac{1}{x-2}$$ is a problem when x-2 equals 0.

### Eureka Math Algebra 1 Module 1 Lesson 17 Exercise Answer Key

Exercise 1.
Solve each equation for x.
a. x-10=0
{10}

b. $$\frac{x}{2}$$+20=0
{-40}

c. Demanding Dwight insists that you give him two solutions to the following equation:
(x-10)($$\frac{x}{2}$$+20)=0
Can you provide him with two solutions?
{10,-40}

d. Demanding Dwight now wants FIVE solutions to the following equation:
(x-10)(2x+6)(x2-36)(x2+10)($$\frac{x}{2}$$+20)=0
Can you provide him with five solutions?
{-40,-6,-3,6,10}

Do you think there might be a sixth solution?
There are exactly 5 solutions.

Exercise 1 (continued)
Give students a few minutes to complete parts (e) and (f), and elicit student responses.

Consider the equation (x-4)(x+3)=0.

e. Rewrite the equation as a compound statement.
x-4=0 or x+3=0

f. Find the two solutions to the equation.
{-3,4}

Exercises 2–7
Give students time to work on the problems individually. As students finish, have them work the problems on the board. Answers are below.

Exercises 2–7

Exercise 2.
(x+1)(x+2)=0
{-2,-1}

Exercise 3.
(3x-2)(x+12)=0
{-12,$$\frac{2}{3}$$}

Exercise 4.
(x-3)(x-3)=0
{3}

Exercise 5.
(x+4)(x-6)(x-10)=0
{-4,6,10}

Exercise 6.
x2-6x=0
{0,6}

Exercise 7.
x(x-5)+4(x-5)=0
{-4,5}

Exercises 8–11
Give students time to work on Exercises 8–11 in pairs. Then, elicit student responses. Remind students of the danger of multiplying both sides by a variable expression.

Exercises 8–11

Exercise 8.
Use factoring to solve the equation for x: (x-2)(2x-3)=(x-2)(x+1).
{2,4}

Exercise 9.
Solve each of the following for x:
a. x+2=5
{3}

b. x2+2x=5x
{0,3}

c. x(5x-20)+2(5x-20)=5(5x-20)
{3,4}

Exercise 10.
Verify: (a-5)(a+5)=a2-25.
a2+5a-5a-25=a2-25
a2-25=a2-25

b. Verify: (x-88)(x+88)= x2-882.
x2+88x-88x-882=x2-882
x2-882=x2-882

c. Verify: A2-B2=(A-B)(A+B).
A2-B2=A2+AB-AB-B2
A2-B2=A2-B2

d. Solve for x: x2-9=5(x-3).
{2,3}

e. Solve for w: (w+2)(w-5)=w2-4.
{-2}

Exercise 11.
A string 60 inches long is to be laid out on a tabletop to make a rectangle of perimeter 60 inches. Write the width of the rectangle as 15+x inches. What is an expression for its length? What is an expression for its area? What value for x gives an area of the largest possible value? Describe the shape of the rectangle for this special value of x.
Length: 15-x Area: (15-x)(15+x)
The largest area is when x=0. In this case, the rectangle is a square with length and width both equal to 15.

### Eureka Math Algebra 1 Module 1 Lesson 17 Problem Set Answer Key

Question 1.
Find the solution set of each equation:
a. (x-1)(x-2)(x-3)=0
{1,2,3}

b. (x-16.5)(x-109)=0
{16.5,109}

c. x(x+7)+5(x+7)=0
{-7,-5}

d. x2+8x+15=0
{-5,-3}

e. (x-3)(x+3)=8x
{-1,9}

Question 2.
Solve x2-11x=0, for x.
{0,11}

Question 3.
Solve (p+3)(p-5)=2(p+3), for p. What solution do you lose if you simply divide by p+3 to get p-5=2?
p=-3 or p=7. The lost solution is p=-3. We assumed p+3 was not zero when we divided by p+3; therefore, our solution was only complete for p values not equal to -3.

Question 4.
The square of a number plus 3 times the number is equal to 4. What is the number?
Solve x2+3x=4, for x, to obtain x=-4 or x=1.

Question 5.
In the right triangle shown below, the length of side AB is x, the length of side BC is x+2, and the length of the hypotenuse AC is x+4. Use this information to find the length of each side. (Use the Pythagorean theorem to get an equation, and solve for x.)
Use the Pythagorean theorem to get the equation x2+(x+2)2=(x+4)2. This is equivalent to x2-4x-12=0, and the solutions are -2 and 6. Choose 6 since x represents a length, and the lengths are

AB: 6
BC: 8
AC: 10

Question 6.
Using what you learned in this lesson, create an equation that has 53 and 22 as its only solutions.
(x-22)(x-53)=0

### Eureka Math Algebra 1 Module 1 Lesson 17 Exit Ticket Answer Key

Question 1.
Find the solution set to the equation 3x2 + 27x = 0.
3x(x+9)=0
3x=0 or x+9=0
x=0 or x=-9
Solution set: {-9,0}

Question 2.
Determine if each statement is true or false. If the statement is false, explain why, or show work proving that it is false.
a. If a=5, then ac=5c.