## Engage NY Eureka Math 8th Grade Module 7 Lesson 5 Answer Key

### Eureka Math Grade 8 Module 7 Lesson 5 Example Answer Key

Example 1.

x^{3} + 9x = \(\frac{1}{2}\) (18x + 54)

Answer:

Now that we know about square roots and cube roots, we can combine that knowledge with our knowledge of the properties of equality to begin solving nonlinear equations like x^{3} + 9x = \(\frac{1}{2}\) (18x + 54). Transform the equation until you can determine the positive value of x that makes the equation true.

Challenge students to solve the equation independently or in pairs. Have students share their strategy for solving the equation. Ask them to explain each step.

x^{3} + 9x = \(\frac{1}{2}\) (18x + 54)

x^{3} + 9x = 9x + 27

x^{3} + 9x – 9x = 9x – 9x + 27

x^{3} = 27

\(\sqrt[3]{x^{3}}\) = \(\sqrt [ 3 ]{ 27 }\)

x = \(\sqrt[3]{3^{3}}\)

x = 3

Now, we verify our solution is correct.

3^{3} + 9(3) = \(\frac{1}{2}\) (18(3) + 54)

27 + 27 = \(\frac{1}{2}\) (54 + 54)

54 = \(\frac{1}{2}\) (108)

54 = 54

Since the left side is the same as the right side, our solution is correct.

Example 2.

x(x – 3) – 51 = – 3x + 13

Answer:

Letâ€™s look at another nonlinear equation. Find the positive value of x that makes the equation true: x(x – 3) – 51 = – 3x + 13.

Provide students with time to solve the equation independently or in pairs. Have students share their strategy for solving the equation. Ask them to explain each step.

Sample response:

x(x – 3) – 51 = – 3x + 13

x^{2} – 3x – 51 = – 3x + 13

x^{2} – 3x + 3x – 51 = – 3x + 3x + 13

x^{2} – 51 = 13

x^{2} – 51 + 51 = 13 + 51

x^{2} = 64

\(\sqrt{x^{2}}\) = Â±\(\sqrt{64}\)

x = Â±\(\sqrt{64}\)

x = Â±8

Now we verify our solution is correct.

Provide students time to check their work.

Let x = 8.

8(8 – 3) – 51 = – 3(8) + 13

8(5) – 51 = – 24 + 13

40 – 51 = – 11

– 11 = – 11

Let x = – 8.

– 8( – 8 – 3) – 51 = – 3( – 8) + 13

– 8( – 11) – 51 = 24 + 13

88 – 51 = 37

37 = 37

Now it is clear that the left side is exactly the same as the right side, and our solution is correct.

### Eureka Math Grade 8 Module 7 Lesson 5 Exercise Answer Key

Find the positive value of x that makes each equation true, and then verify your solution is correct.

Exercise 1.

a. Solve x^{2} – 14 = 5x + 67 – 5x.

Answer:

x^{2} – 14 = 5x + 67 – 5x

x^{2} – 14 = 67

x^{2} – 14 + 14 = 67 + 14

x^{2} = 81

\(\sqrt{x^{2}}\) = Â±\(\sqrt{81}\)

x = Â±\(\sqrt{81}\)

x = Â±9

Check:

9^{2} – 14 = 5(9) + 67 – 5(9)

81 – 14 = 45 + 67 – 45

67 = 67

( – 9)^{2} – 14 = 5( – 9) + 67 – 5( – 9)

81 – 14 = – 45 + 67 + 45

67 = 67

b. Explain how you solved the equation.

Answer:

To solve the equation, I had to first use the properties of equality to transform the equation into the form of x^{2} = 81. Then, I had to take the square root of both sides of the equation to determine that x = 9 since the number x is being squared.

Exercise 2.

Solve and simplify: x(x – 1) = 121 – x.

Answer:

x(x – 1) = 121 – x

x^{2} – x = 121 – x

x^{2} – x + x = 121 – x + x

x^{2} = 121

\(\sqrt{x^{2}}\) = Â±\(\sqrt{121}\)

x = Â±\(\sqrt{121}\)

x = Â±11

Check:

11(11 – 1) = 121 – 11

11(10) = 110

110 = 110

– 11( – 11 – 1) = 121 – ( – 11)

– 11( – 12) = 121 + 11

132 = 132

Exercise 3.

A square has a side length of 3x inches and an area of 324 in^{2}. What is the value of x?

Answer:

(3x)^{2} = 324

3^{2} x^{2} = 324

9x^{2} = 324

\(\frac{9 x^{2}}{9}\) = \(\frac{324}{9}\)

x^{2} = 36

\(\sqrt{x^{2}}\) = \(\sqrt{36}\)

x = 6

Check:

(3(6)) ^{2} = 324

18^{2} = 324

324 = 324

A negative number would not make sense as a length, so x = 6.

Exercise 4.

– 3x^{3} + 14 = – 67

Answer:

– 3x^{3} + 14 = – 67

– 3x^{3} + 14 – 14 = – 67 – 14

– 3x^{3} = – 81

\(\frac{ – 3 x^{3}}{3}\) = \(\frac{ – 81}{ – 3}\)

x^{3} = 27

\(\sqrt[3]{x^{3}}\) = \(\sqrt [ 3 ]{ 27 }\)

x = 3

Check:

– 3(3)^{3} + 14 = – 67

– 3(27) + 14 = – 67

– 81 + 14 = – 67

– 67 = – 67

Exercise 5.

x(x + 4) – 3 = 4(x + 19.5)

Answer:

x(x + 4) – 3 = 4(x + 19.5)

x^{2} + 4x – 3 = 4x + 78

x^{2} + 4x – 4x – 3 = 4x – 4x + 78

x^{2} – 3 = 78

x^{2} – 3 + 3 = 78 + 3

x^{2} = 81

\(\sqrt{x^{2}}\) = Â±\(\sqrt{81}\)

x = Â±9

Check:

9(9 + 4) – 3 = 4(9 + 19.5)

9(13) – 3 = 4(28.5)

117 – 3 = 114

114 = 114

– 9( – 9 + 4) – 3 = 4( – 9 + 19.5)

– 9( – 5) – 3 = 4(10.5)

45 – 3 = 42

42 = 42

Exercise 6.

216 + x = x(x^{2} – 5) + 6x

Answer:

216 + x = x(x^{2} – 5) + 6x

216 + x = x^{3} – 5x + 6x

216 + x = x^{3} + x

216 + x – x = x^{3} + x – x

216 = x^{3}

\(\sqrt [ 3 ]{ 216 }\) = \(\sqrt[3]{x^{3}}\)

6 = x

Check:

216 + 6 = 6(6^{2} – 5) + 6(6)

222 = 6(31) + 36

222 = 186 + 36

222 = 222

Exercise 7.

a. What are we trying to determine in the diagram below?

Answer:

We need to determine the value of x so that its square root, multiplied by 4, satisfies the equation

5^{2} + (4\(\sqrt{x}\))^{2} = 11^{2}.

b. Determine the value of x, and check your answer.

Answer:

5^{2} + (4\(\sqrt{x}\))^{2} = 11^{2}

25 + 4^{2} (\(\sqrt{x}\))^{2} = 121

25 – 25 + 4^{2} (\(\sqrt{x}\))^{2} = 121 – 25

16x = 96

\(\frac{16x}{16}\) = \(\frac{96}{16}\)

x = 6

The value of x is 6.

Check:

5^{2} + (4\(\sqrt{6}\))^{2} = 11^{2}

25 + 16(6) = 121

25 + 96 = 121

121 = 121

### Eureka Math Grade 8 Module 7 Lesson 5 Problem Set Answer Key

Find the positive value of x that makes each equation true, and then verify your solution is correct.

Question 1.

x^{2} (x + 7) = \(\frac{1}{2}\) (14x^{2} + 16)

Answer:

x^{2} (x + 7) = \(\frac{1}{2}\) (14x^{2} + 16)

x^{3} + 7x^{2} = 7x^{2} + 8

x^{3} + 7x^{2} – 7x^{2} = 7x^{2} – 7x^{2} + 8

x^{3} = 8

\(\sqrt[3]{x^{3}}\) = \(\sqrt [ 3 ]{ 8 }\)

x = 2

Check:

2^{2} (2 + 7) = \(\frac{1}{2}\) (14(2^{2} ) + 16)

4(9) = \(\frac{1}{2}\) (56 + 16)

36 = \(\frac{1}{2}\) (72)

36 = 36

Question 2.

x^{3} = 1331^{ – 1}

Answer:

x^{3} = 1331^{ – 1}

\(\sqrt[3]{x^{3}}\) = \(\sqrt[3]{1331^{ – 1}}\)

x = \(\sqrt[3]{\frac{1}{1331}}\)

x = \(\sqrt[3]{\frac{1}{11^{3}}}\)

x = \(\frac{1}{11}\)

Check:

(\(\left(\frac{1}{11}\right)^{3}\))^{3} = 1331^{ – 1}

\(\frac{1}{11^{3}}\) = 1331^{ – 1}

\(\frac{1}{1331}\) = 1331^{ – 1}

1331^{ – 1} = 1331^{ – 1}

Question 3.

Determine the positive value of x that makes the equation true, and then explain how you solved the equation.

\(\frac{x^{9}}{x^{7}}\) – 49 = 0

Answer:

\(\frac{x^{9}}{x^{7}}\) – 49 = 0

x^{2} – 49 = 0

x^{2} – 49 + 49 = 0 + 49

x^{2} = 49

\(\sqrt{x^{2}}\) = \(\sqrt{49}\)

x = 7

Check:

7^{2} – 49 = 0

49 – 49 = 0

0 = 0

To solve the equation, I first had to simplify the expression \(\frac{x^{9}}{x^{7}}\) to x^{2}. Next, I used the properties of equality to transform the equation into x^{2} = 49. Finally, I had to take the square root of both sides of the equation to solve for x.

Question 4.

Determine the positive value of x that makes the equation true.

(8x)^{2} = 1

Answer:

(8x)^{2} = 1

64x^{2} = 1

\(\sqrt{64^{2}}\) = \(\sqrt{1}\)

8x = 1

\(\frac{8x}{8}\) = \(\frac{1}{8}\)

x = \(\frac{1}{8}\)

Check:

(8(\(\frac{1}{8}\)))^{2} = 1

1^{2} = 1

1 = 1

Question 5.

(9\(\sqrt{x}\))^{2} – 43x = 76

Answer:

(9\(\sqrt{x}\))^{2} – 43x = 76

9^{2} (âˆšx)^{2} – 43x = 76

81x – 43x = 76

38x = 76

\(\frac{38x}{38}\) = \(\frac{76}{38}\)

x = 2

Check:

(9(\(\sqrt{2}\)))^{2} – 43(2) = 76

9^{2} (\(\sqrt{2}\))^{2} – 86 = 76

81(2) – 86 = 76

162 – 86 = 76

76 = 76

Question 6.

Determine the length of the hypotenuse of the right triangle below.

Answer:

3^{2} + 7^{2} = x^{2}

9 + 49 = x^{2}

58 = x^{2}

\(\sqrt{58}\) = \(\sqrt{x^{2}}\)

\(\sqrt{58}\) = x

Check:

3^{2} + 7^{2} = (\(\sqrt{52}\))^{2}

9 + 49 = 58

58 = 58

Since x = \(\sqrt{58}\), the length of the hypotenuse is \(\sqrt{58}\) mm.

Question 7.

Determine the length of the legs in the right triangle below.

Answer:

x^{2} + x^{2} = (14\(\sqrt{2}\))^{2}

2x^{2} = 14^{2} (\(\sqrt{2}\))^{2}

2x^{2} = 196(2)

\(\frac{2 x^{2}}{2}\) = \(\frac{196(2)}{2}\)

x^{2} = 196

\(\sqrt{x^{2}}\) = \(\sqrt{196}\)

x = \(\sqrt{14^{2}}\)

x = 14

Check:

14^{2} + 14^{2} = (14\(\sqrt{2}\))^{2}

196 + 196 = 14^{2} (\(\sqrt{2}\))^{2}

392 = 196(2)

392 = 392

Since x = 14, the length of each of the legs of the right triangle is 14 cm.

Question 8.

An equilateral triangle has side lengths of 6 cm. What is the height of the triangle? What is the area of the triangle?

Answer:

Note: This problem has two solutions, one with a simplified root and one without. Choose the appropriate solution for your classes based on how much simplifying you have taught them.

Let h cm represent the height of the triangle.

3^{2} + h^{2} = 6^{2}

9 + h^{2} = 36

9 – 9 + h^{2} = 36 – 9

h^{2} = 27

\(\sqrt{h^{2}}\) = \(\sqrt{27}\)

h = \(\sqrt{27}\)

h = \(\sqrt{3^{3}}\)

h = \(\sqrt{3^{2}}\)Ã—\(\sqrt{3}\)

h = 3\(\sqrt{3}\)

Let A represent the area of the triangle.

A = \(\frac{6(3 \sqrt{3})}{2}\))

A = 3(3\(\sqrt{3}\))

A = 9\(\sqrt{3}\)

Simplified: The height of the triangle is 3\(\sqrt{3}\) cm, and the area is 9\(\sqrt{3}\) cm^{2}.

Unsimplified: The height of the triangle is \(\sqrt{27}\) cm, and the area is 3\(\sqrt{27}\) cm^{2}

Question 9.

Challenge: Find the positive value of x that makes the equation true.

(\(\frac{1}{2}\) x)^{2} – 3x = 7x + 8 – 10x

Answer:

(\(\frac{1}{2}\) x)^{2} – 3x = 7x + 8 – 10x

\(\frac{1}{4}\) x^{2} – 3x = – 3x + 8

\(\frac{1}{4}\) x^{2} – 3x + 3x = – 3x + 3x + 8

\(\frac{1}{4}\) x^{2} = 8

4(\(\frac{1}{4}\)) x^{2} = 8(4)

x^{2} = 32

\(\sqrt{x^{2}}\) = \(\sqrt{32}\)

x = \(\sqrt{2^{5}}\)

x = \(\sqrt{2^{2}}\) â‹… \(\sqrt{2^{2}}\) â‹… \(\sqrt{2}\)

x = 4\(\sqrt{2}\)

Check:

(\(\frac{1}{2}\) (4\(\sqrt{2}\)))^{2} – 3(4\(\sqrt{2}\)) = 7(4\(\sqrt{2}\)) + 8 – 10(4\(\sqrt{2}\))

\(\frac{1}{4}\) (16)(2) – 3(4\(\sqrt{2}\)) = 7(4\(\sqrt{2}\)) – 10(4\(\sqrt{2}\)) + 8

\(\frac{32}{4}\) – 3(4\(\sqrt{2}\)) = 7(4\(\sqrt{2}\)) – 10(4\(\sqrt{2}\)) + 8

8 – 3(4\(\sqrt{2}\)) = (7 – 10)(4\(\sqrt{2}\)) + 8

8 – 3(4\(\sqrt{2}\)) = – 3(4\(\sqrt{2}\)) + 8

8 – 8 – 3(4\(\sqrt{2}\)) = – 3(4\(\sqrt{2}\)) + 8 – 8

– 3(4\(\sqrt{2}\)) = – 3(4\(\sqrt{2}\))

Question 10.

Challenge: Find the positive value of x that makes the equation true.

11x + x(x – 4) = 7(x + 9)

Answer:

11x + x(x – 4) = 7(x + 9)

11x + x^{2} – 4x = 7x + 63

7x + x^{2} = 7x + 63

7x – 7x + x^{2} = 7x – 7x + 63

x^{2} = 63

\(\sqrt{x^{2}}\)) = \(\sqrt{63}\)

x = \(\sqrt{\left(3^{2}\right)(7)}\)

x = \(\sqrt{3^{2}}\) â‹… \(\sqrt{7}\)

x = 3\(\sqrt{7}\)

Check:

11(3\(\sqrt{7}\)) + 3\(\sqrt{7}\) (3\(\sqrt{7}\) – 4) = 7(3\(\sqrt{7}\) + 9)

33\(\sqrt{7}\) + 3^{2} (\(\sqrt{7}\))^{2} – 4(3\(\sqrt{7}\)) = 21\(\sqrt{7}\) + 63

33\(\sqrt{7}\) – 4(3\(\sqrt{7}\)) + 9(7) = 21\(\sqrt{7}\) + 63

33\(\sqrt{7}\) – 12\(\sqrt{7}\) + 63 = 21\(\sqrt{7}\) + 63

(33 – 12) \(\sqrt{7}\) + 63 = 21\(\sqrt{7}\) + 63

21\(\sqrt{7}\) + 63 = 21\(\sqrt{7}\) + 63

21\(\sqrt{7}\) + 63 – 63 = 21\(\sqrt{7}\) + 63 – 63

21\(\sqrt{7}\) = 21\(\sqrt{7}\)

### Eureka Math Grade 8 Module 7 Lesson 5 Exit Ticket Answer Key

Question 1.

Find the positive value of x that makes the equation true, and then verify your solution is correct.

x^{2} + 4x = 4(x + 16)

Answer:

x^{2} + 4x = 4(x + 16)

x^{2} + 4x = 4x + 64

x^{2} + 4x – 4x = 4x – 4x + 64

x^{2} = 64

\(\sqrt{x^{2}}\) = \(\sqrt{64}\)

x = 8

Check:

8^{2} + 4(8) = 4(8 + 16)

64 + 32 = 4(24)

96 = 96

Question 2.

Find the positive value of x that makes the equation true, and then verify your solution is correct.

(4x)^{3} = 1728

Answer:

(4x)^{3} = 1728

64x^{3} = 1728

\(\frac{1}{64}\)(64x^{3}) = (1728)\(\frac{1}{64}\)

x^{3} = 27

\(\sqrt[3]{x^{3}}\) = \(\sqrt [ 3 ]{ 27 }\)

x = 3

Check:

(4(3))^{3} = 1728

12^{3} = 1728

1728 = 1728