## 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 2x^{2}-10x=0, for x.

Answer:

{0,5}

Example 2.

Solve x(x-3)+5(x-3)= 0, for x.

Answer:

{-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?

Answer:

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.

Answer:

(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

Answer:

{10}

b. \(\frac{x}{2}\)+20=0

Answer:

{-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?

Answer:

{10,-40}

d. Demanding Dwight now wants FIVE solutions to the following equation:

(x-10)(2x+6)(x^{2}-36)(x^{2}+10)(\(\frac{x}{2}\)+20)=0

Can you provide him with five solutions?

Answer:

{-40,-6,-3,6,10}

Do you think there might be a sixth solution?

Answer:

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.

Answer:

x-4=0 or x+3=0

f. Find the two solutions to the equation.

Answer:

{-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

Answer:

{-2,-1}

Exercise 3.

(3x-2)(x+12)=0

Answer:

{-12,\(\frac{2}{3}\)}

Exercise 4.

(x-3)(x-3)=0

Answer:

{3}

Exercise 5.

(x+4)(x-6)(x-10)=0

Answer:

{-4,6,10}

Exercise 6.

x^{2}-6x=0

Answer:

{0,6}

Exercise 7.

x(x-5)+4(x-5)=0

Answer:

{-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).

Answer:

{2,4}

Exercise 9.

Solve each of the following for x:

a. x+2=5

Answer:

{3}

b. x^{2}+2x=5x

Answer:

{0,3}

c. x(5x-20)+2(5x-20)=5(5x-20)

Answer:

{3,4}

Exercise 10.

Verify: (a-5)(a+5)=a^{2}-25.

Answer:

a^{2}+5a-5a-25=a^{2}-25

a^{2}-25=a^{2}-25

b. Verify: (x-88)(x+88)= x^{2}-88^{2}.

Answer:

x^{2}+88x-88x-88^{2}=x^{2}-88^{2}

x^{2}-88^{2}=x^{2}-88^{2}

c. Verify: A^{2}-B^{2}=(A-B)(A+B).

Answer:

A^{2}-B^{2}=A^{2}+AB-AB-B^{2}

A^{2}-B^{2}=A^{2}-B^{2}

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

Answer:

{2,3}

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

Answer:

{-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.

Answer:

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

Answer:

{1,2,3}

b. (x-16.5)(x-109)=0

Answer:

{16.5,109}

c. x(x+7)+5(x+7)=0

Answer:

{-7,-5}

d. x^{2}+8x+15=0

Answer:

{-5,-3}

e. (x-3)(x+3)=8x

Answer:

{-1,9}

Question 2.

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

Answer:

{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?

Answer:

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?

Answer:

Solve x^{2}+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.)

Answer:

Use the Pythagorean theorem to get the equation x^{2}+(x+2)^{2}=(x+4)^{2}. This is equivalent to x^{2}-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.

Answer:

(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 3x^{2} + 27x = 0.

Answer:

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.

Answer:

True.

b. If ac=5c, then a=5.

Answer:

False. a could equal 5, or c could equal 0.

ac=5c

ac-5c=0

c(a-5)=0

c=0 or a-5=0

c=0 or a=5