## Engage NY Eureka Math 8th Grade Module 7 Lesson 4 Answer Key

### Eureka Math Grade 8 Module 7 Lesson 4 Example Answer Key

Example 1.

Simplify the square root as much as possible.

\(\sqrt{50}\) =

Answer:

→ Is the number 50 a perfect square? Explain.

The number 50 is not a perfect square because there is no integer squared that equals 50.

→ Since 50 is not a perfect square, when we need to simplify \(\sqrt{50}\), we write the factors of the number 50 looking specifically for those that are perfect squares. What are the factors of 50?

50 = 2 × 5^{2}

Since 50 = 2 × 5^{2}, then \(\sqrt{50}\) = \(\sqrt{2 \times 5^{2}}\). We can rewrite \(\sqrt{50}\) as a product of its factors:

\(\sqrt{50}\) = \(\sqrt{2}\) × \(\sqrt{5^{2}}\).

→ Obviously, 5^{2} is a perfect square. Therefore, \(\sqrt{5^{2}}\) = 5, so \(\sqrt{50}\) = 5 × \(\sqrt{2}\) = 5\(\sqrt{2}\). Since \(\sqrt{2}\) is not a perfect square, we leave it as it is. We have simplified this expression as much as possible because there are no other perfect square factors remaining in the square root.

→ The number \(\sqrt{50}\) is said to be in its simplified form when all perfect square factors have been simplified. Therefore, 5\(\sqrt{2}\) is the simplified form of \(\sqrt{50}\).

Now that we know \(\sqrt{50}\) can be expressed as a product of its factors, we also know that we can multiply expressions containing square roots. For example, if we are given \(\sqrt{2}\) × \(\sqrt{5^{2}}\), we can rewrite the expression as √(2 × 5^{2} ) = \(\sqrt{50}\).

Example 2.

Simplify the square root as much as possible.

\(\sqrt{28}\) =

Answer:

→ Is the number 28 a perfect square? Explain.

The number 28 is not a perfect square because there is no integer squared that equals 28.

→ What are the factors of 28?

28 = 2^{2} × 7

Since 28 = 2^{2} × 7, then \(\sqrt{28}\) = \(\sqrt{2^{2} \times 7}\). We can rewrite \(\sqrt{28}\) as a product of its factors:

\(\sqrt{28}\) = \(\sqrt{2^{2}}\) × \(\sqrt{7}\).

→ Obviously, 2^{2} is a perfect square. Therefore, \(\sqrt{2^{2}}\) = 2, and \(\sqrt{28}\) = 2 × \(\sqrt{7}\) = 2\(\sqrt{7}\). Since \(\sqrt{7}\) is not a perfect square, we leave it as it is.

→ The number \(\sqrt{28}\) is said to be in its simplified form when all perfect square factors have been simplified. Therefore, 2\(\sqrt{7}\) is the simplified form of \(\sqrt{28}\).

Example 3.

Simplify the square root as much as possible.

\(\sqrt{128}\) =

Answer:

→ In this example, students may or may not recognize 128 as 64 × 2. The work below assumes that they do not. Consider showing students the solution below, as well as this alternative solution:

\(\sqrt{128}\) = \(\sqrt{64 \times 2}\) = \(\sqrt{64}\) × \(\sqrt{2}\) = 8 × \(\sqrt{2}\) = 8\(\sqrt{2}\).

→ Is the number 128 a perfect square? Explain.

The number 128 is not a perfect square because there is no integer squared that equals 128.

→ What are the factors of 128?

128 = 2^{7}

→ Since 128 = 2^{7}, then \(\sqrt{128}\) = √(2^{7} ). We know that we can simplify perfect squares, so we can rewrite 2^{7} as 2^{2} × 2^{2} × 2^{2} × 2 because of what we know about the laws of exponents. Then,

\(\sqrt{128}\) = \(\sqrt{2^{2}}\) × \(\sqrt{2^{2}}\) × \(\sqrt{2^{2}}\) × \(\sqrt{2}\).

Each 2^{2} is a perfect square. Therefore, \(\sqrt{128}\) = 2 × 2 × 2 × \(\sqrt{2}\) = 8\(\sqrt{2}\).

Example 4.

Simplify the square root as much as possible.

\(\sqrt{288}\) =

Answer:

In this example, students may or may not recognize 288 as 144 × 2. The work below assumes that they do not. Consider showing students the solution below, as well as this alternative solution:

\(\sqrt{288}\) = \(\sqrt{144 \times 2}\) = \(\sqrt{144}\) × \(\sqrt{2}\) = 12 × \(\sqrt{2}\) = 12\(\sqrt{2}\).

→ Is the number 288 a perfect square? Explain.

The number 288 is not a perfect square because there is no integer squared that equals 288.

→ What are the factors of 288?

288 = 2^{5} × 3^{2}

Since 288 = 2^{5} × 3^{2}, then \(\sqrt{288}\) = √(2^{5} × 3^{2} ). What do we do next?

Use the laws of exponents to rewrite 2^{5} as 2^{2} × 2^{2} × 2.

→ Then, \(\sqrt{288}\) is equivalent to

\(\sqrt{288}\) = \(\sqrt{2^{2}}\) × \(\sqrt{2^{2}}\) × \(\sqrt{2}\) × \(\sqrt{3^{2}}\).

→ What does this simplify to?

\(\sqrt{288}\) = \(\sqrt{2^{2}}\) × \(\sqrt{2^{2}}\) × \(\sqrt{2}\) × \(\sqrt{3^{2}}\)) = \(\sqrt{2^{2}}\) × \(\sqrt{2^{2}}\) × \(\sqrt{3^{2}}\) ) × \(\sqrt{2}\) = 2 × 2 × 3 × \(\sqrt{2}\) = 12\(\sqrt{2}\)

### Eureka Math Grade 8 Module 7 Lesson 4 Exercise Answer Key

Opening Exercise

a.

i. What does \(\sqrt{16}\) equal?

Answer:

4

ii. What does 4 × 4 equal?

Answer:

16

iii. Does \(\sqrt{16}\) = \(\sqrt{4 \times 4}\)?

Answer:

Yes

b.

i. What does \(\sqrt{36}\) equal?

Answer:

6

ii. What does 6 × 6 equal?

Answer:

36

iii. Does \(\sqrt{36}\) = \(\sqrt{6 \times 6}\)?

Answer:

Yes

c.

i. What does \(\sqrt{121}\) equal?

Answer:

11

ii. What does 11 × 11 equal?

Answer:

121

iii. Does \(\sqrt{121}\) = \(\sqrt{11 \times 11}\)?

Yes

d.

i. What does \(\sqrt{81}\) equal?

Answer:

9

ii. What does 9 × 9 equal?

Answer:

81

iii. Does \(\sqrt{81}\) = \(\sqrt{9 \times 9}\)?

Answer:

Yes

e. Rewrite \(\sqrt{20}\) using at least one perfect square factor.

Answer:

\(\sqrt{20}\) = \(\sqrt{4 \times 5}\)

f. Rewrite \(\sqrt{28}\) using at least one perfect square factor.

Answer:

\(\sqrt{28}\) = \(\sqrt{4 \times 7}\)

Exercises 1–4

Simplify the square roots as much as possible.

Exercise 1.

\(\sqrt{18}\)

Answer:

\(\sqrt{18}\) = \(\sqrt{2 \times 3^{2}}\)

= \(\sqrt{2}\) × \(\sqrt{3^{2}}\)

= 3\(\sqrt{2}\)

Exercise 2.

\(\sqrt{44}\)

Answer:

\(\sqrt{44}\) = \(\sqrt{2^{2} \times 11}\)

= \(\sqrt{2^{2}}\) × \(\sqrt{11}\)

= 2\(\sqrt{11}\)

Exercise 3.

\(\sqrt{169}\)

Answer:

\(\sqrt{169}\) = \(\sqrt{13^{2}}\)

= 13

Exercise 4.

\(\sqrt{75}\)

Answer:

\(\sqrt{75}\) = \(\sqrt{3 \times 5^{2}}\)

= \(\sqrt{3}\) × \(\sqrt{5^{2}}\)

= 5\(\sqrt{3}\)

Exercises 5–8

Exercise 5.

Simplify \(\sqrt{108}\).

Answer:

\(\sqrt{108}\) = \(\sqrt{2^{2} \times 3^{3}}\)

= \(\sqrt{2^{2}}\) × \(\sqrt{3^{2}}\) × \(\sqrt{3}\)

= 2 × 3\(\sqrt{3}\)

= 6\(\sqrt{3}\)

Exercise 6.

Simplify \(\sqrt{250}\).

Answer:

\(\sqrt{250}\) = \(\sqrt{2 \times 5^{3}}\)

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

= 5\(\sqrt{2}\) × \(\sqrt{5}\)

= 5\(\sqrt{10}\)

Exercise 7.

Simplify \(\sqrt{200}\).

Answer:

\(\sqrt{200}\) = \(\sqrt{2^{3} \times 5^{2}}\)

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

= 2 × 5\(\sqrt{2}\)

= 10\(\sqrt{2}\)

Exercise 8.

Simplify \(\sqrt{504}\).

Answer:

\(\sqrt{504}\) = \(\sqrt{2^{3} \times 3^{2} \times 7}\)

= \(\sqrt{2^{2}}\) × \(\sqrt{2}\) × \(\sqrt{3^{2}}\) × \(\sqrt{7}\)

= 2 × 3 × \(\sqrt{2}\) × \(\sqrt{7}\)

= 6\(\sqrt{14}\)

### Eureka Math Grade 8 Module 7 Lesson 4 Problem Set Answer Key

Simplify each of the square roots in Problems 1–5 as much as possible.

Question 1.

\(\sqrt{98}\)

\(\sqrt{98}\) = \(\sqrt{2 \times 7^{2}}\)

= \(\sqrt{2}\) × \(\sqrt{7^{2}}\)

= 7\(\sqrt{2}\)

Question 2.

\(\sqrt{54}\)

\(\sqrt{54}\) = \(\sqrt{2 \times 3^{3}}\)

= \(\sqrt{2}\) × \(\sqrt{3}\) × \(\sqrt{3^{2}}\)

= 3\(\sqrt{6}\)

Question 3.

\(\sqrt{144}\)

\(\sqrt{144}\) = \(\sqrt{12^{2}}\)

= 12

Question 4.

\(\sqrt{512}\)

\(\sqrt{512}\) = \(\sqrt{2^{9}}\)

= \(\sqrt{2^{2}}\) × \(\sqrt{2^{2}}\) × \(\sqrt{2^{2}}\) × \(\sqrt{2^{2}}\) × \(\sqrt{2}\)

= 2 × 2 × 2 × 2\(\sqrt{2}\)

= 16\(\sqrt{2}\)

Question 5.

\(\sqrt{756}\)

\(\sqrt{756}\) = \(\sqrt{2^{2} \times 3^{3} \times 7}\)

= \(\sqrt{2^{2}}\) × \(\sqrt{3^{2}}\) × \(\sqrt{3}\) × \(\sqrt{7}\)

= 2 × 3 × \(\sqrt{21}\)

= 6\(\sqrt{21}\)

Question 6.

What is the length of the unknown side of the right triangle? Simplify your answer, if possible.

Answer:

Let c units represent the length of the hypotenuse.

(\(\sqrt{27}\))^{2 }+ (\(\sqrt{48}\))^{2} = c^{2}

27 + 48 = c^{2}

75 = c^{2}

\(\sqrt{75}\) = \(\sqrt{c^{2}}\)

\(\sqrt{5^{2}}\) × \(\sqrt{3}\) = c

5\(\sqrt{3}\) = c

The length of the hypotenuse is 5\(\sqrt{3}\) units.

Question 7.

What is the length of the unknown side of the right triangle? Simplify your answer, if possible.

Answer:

Let c cm represent the length of the hypotenuse.

3^{2 }+ 8^{2} = c^{2}

9 + 64 = c^{2}

73 = c^{2}

\(\sqrt{73}\) = \(\sqrt{c^{2}}\)

\(\sqrt{73}\) = c

The length of the unknown side is \(\sqrt{73}\) cm.

Question 8.

What is the length of the unknown side of the right triangle? Simplify your answer, if possible.

Answer:

Let c mm represent the length of the hypotenuse.

3^{2 }+ 3^{2} = c^{2}

9 + 9 = c^{2}

18 = c^{2}

\(\sqrt{18}\) = \(\sqrt{c^{2}}\)

\(\sqrt{18}\) = c

\(\sqrt{3^{2}}\) × \(\sqrt{2}\) = c

3\(\sqrt{2}\) = c

The length of the unknown side is 3\(\sqrt{2}\) mm.

Question 9.

What is the length of the unknown side of the right triangle? Simplify your answer, if possible.

Answer:

Let x in. represent the unknown length.

x^{2 }+ 8^{2} = 12^{2}

x^{2 }+ 64 = 144

x^{2 }+ 64 – 64 = 144 – 64

x^{2} = 80

\(\sqrt{x^{2}}\) = \(\sqrt{80}\)

x = \(\sqrt{80}\)

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

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

x = 2 ⋅ 2\(\sqrt{5}\)

x = 4\(\sqrt{5}\)

The length of the unknown side is 4\(\sqrt{5}\) in.

Question 10.

Josue simplified \(\sqrt{450}\) as 15\(\sqrt{2}\) Is he correct? Explain why or why not.

Answer:

\(\sqrt{450}\) = \(\sqrt{2 \times 3^{2} \times 5^{2}}\)

= \(\sqrt{2}\) × \(\sqrt{3^{2}}\) × \(\sqrt{5^{2}}\)

= 3 × 5 × \(\sqrt{1}\)

= 15\(\sqrt{1}\)

Yes, Josue is correct because the number 450 = 2 × 3^{2} × 5^{2}. The factors that are perfect squares simplify to 15 leaving just the factor of 2 that cannot be simplified. Therefore, \(\sqrt{450}\) = 15\(\sqrt{2}\).

Question 11.

Tiah was absent from school the day that you learned how to simplify a square root. Using \(\sqrt{360}\), write Tiah an explanation for simplifying square roots.

Answer:

To simplify \(\sqrt{360}\), first write the factors of 360. The number 360 = 2^{3} × 3^{2} × 5. Now, we can use the factors to write \(\sqrt{360}\) = \(\sqrt{2^{3} \times 3^{2} \times 5}\)), which can then be expressed as \(\sqrt{360}\) = \(\sqrt{2^{3}}\) × \(\sqrt{3^{2}}\) × \(\sqrt{5}\). Because we want to simplify square roots, we can rewrite the factor \(\sqrt{2^{3}}\) as \(\sqrt{2^{2}}\) × \(\sqrt{2}\) because of the laws of exponents. Now, we have

\(\sqrt{360}\) = \(\sqrt{2^{2}}\) × √2 × \(\sqrt{3^{2}}\) × \(\sqrt{5}\).

Each perfect square can be simplified as follows:

\(\sqrt{360}\) = 2 × \(\sqrt{2}\) × 3 × \(\sqrt{5}\)

= 2 × 3 × \(\sqrt{2}\) × \(\sqrt{5}\)

= 6\(\sqrt{10}\).

The simplified version of \(\sqrt{360}\) = 6\(\sqrt{10}\).

### Eureka Math Grade 8 Module 7 Lesson 4 Exit Ticket Answer Key

Simplify the square roots as much as possible.

Question 1.

\(\sqrt{24}\)

Answer:

\(\sqrt{24}\) = \(\sqrt{2^{2} \times 6}\)

= \(\sqrt{2^{2}}\) × \(\sqrt{6}\)

= 2\(\sqrt{6}\)

Question 2.

\(\sqrt{338}\)

Answer:

\(\sqrt{338}\) = \(\sqrt{13^{2} \times 2}\)

= \(\sqrt{13^{2}}\) × \(\sqrt{24}\)

= 13\(\sqrt{2}\)

Question 3.

\(\sqrt{196}\)

Answer:

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

= 14

Question 4.

\(\sqrt{2420}\)

Answer:

\(\sqrt{2420}\) = \(\sqrt{2^{2} \times 11^{2} \times 5}\)

= \(\sqrt{2^{2}}\) × \(\sqrt{11^{2}}\) × \(\sqrt{5}\)

= 2 × 11 × \(\sqrt{5}\)

= 22\(\sqrt{5}\)