## 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:

â†’ 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.

â†’ 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 Ã— 5^{2}

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

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

â†’ Obviously, 5^{2} is a perfect square. Therefore, \(\sqrt{5^{2}}\) = 5, so \(\sqrt{50}\) = 5 Ã— \(\sqrt{2}\) = 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.

â†’ 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 \(\sqrt{2}\) Ã— \(\sqrt{5^{2}}\), we can rewrite the expression as âˆš(2 Ã— 5^{2} ) = \(\sqrt{50}\).

Example 2.

Simplify the square root as much as possible.

\(\sqrt{28}\) =

Answer:

â†’ 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.

â†’ What are the factors of 28?

28 = 2^{2} Ã— 7

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

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

â†’ Obviously, 2^{2} is a perfect square. Therefore, \(\sqrt{2^{2}}\) = 2, and \(\sqrt{28}\) = 2 Ã— \(\sqrt{7}\) = 2\(\sqrt{7}\). Since \(\sqrt{7}\) is not a perfect square, we leave it as it is.

â†’ 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:

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

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

â†’ 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.

â†’ What are the factors of 128?

128 = 2^{7}

â†’ Since 128 = 2^{7}, then \(\sqrt{128}\) = âˆš(2^{7} ). We know that we can simplify perfect squares, so we can rewrite 2^{7} as 2^{2} Ã— 2^{2} Ã— 2^{2} Ã— 2 because of what we know about the laws of exponents. Then,

\(\sqrt{128}\) = \(\sqrt{2^{2}}\) Ã— \(\sqrt{2^{2}}\) Ã— \(\sqrt{2^{2}}\) Ã— \(\sqrt{2}\).

Each 2^{2} is a perfect square. Therefore, \(\sqrt{128}\) = 2 Ã— 2 Ã— 2 Ã— \(\sqrt{2}\) = 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 Ã— 2. The work below assumes that they do not. Consider showing students the solution below, as well as this alternative solution:

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

â†’ 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.

â†’ What are the factors of 288?

288 = 2^{5} Ã— 3^{2}

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

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

â†’ Then, \(\sqrt{288}\) is equivalent to

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

â†’ What does this simplify to?

\(\sqrt{288}\) = \(\sqrt{2^{2}}\) Ã— \(\sqrt{2^{2}}\) Ã— \(\sqrt{2}\) Ã— \(\sqrt{3^{2}}\)) = \(\sqrt{2^{2}}\) Ã— \(\sqrt{2^{2}}\) Ã— \(\sqrt{3^{2}}\) ) Ã— \(\sqrt{2}\) = 2 Ã— 2 Ã— 3 Ã— \(\sqrt{2}\) = 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 Ã— 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 Ã— 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 Ã— 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 Ã— 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 1â€“4

Simplify the square roots as much as possible.

Exercise 1.

\(\sqrt{18}\)

Answer:

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

= \(\sqrt{2}\) Ã— \(\sqrt{3^{2}}\)

= 3\(\sqrt{2}\)

Exercise 2.

\(\sqrt{44}\)

Answer:

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

= \(\sqrt{2^{2}}\) Ã— \(\sqrt{11}\)

= 2\(\sqrt{11}\)

Exercise 3.

\(\sqrt{169}\)

Answer:

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

= 13

Exercise 4.

\(\sqrt{75}\)

Answer:

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

= \(\sqrt{3}\) Ã— \(\sqrt{5^{2}}\)

= 5\(\sqrt{3}\)

Exercises 5â€“8

Exercise 5.

Simplify \(\sqrt{108}\).

Answer:

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

= \(\sqrt{2^{2}}\) Ã— \(\sqrt{3^{2}}\) Ã— \(\sqrt{3}\)

= 2 Ã— 3\(\sqrt{3}\)

= 6\(\sqrt{3}\)

Exercise 6.

Simplify \(\sqrt{250}\).

Answer:

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

= \(\sqrt{2}\) Ã— \(\sqrt{5^{2}}\) Ã— \(\sqrt{5}\)

= 5\(\sqrt{2}\) Ã— \(\sqrt{5}\)

= 5\(\sqrt{10}\)

Exercise 7.

Simplify \(\sqrt{200}\).

Answer:

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

= \(\sqrt{2^{2}}\) Ã— \(\sqrt{2}\) Ã— \(\sqrt{5^{2}}\)

= 2 Ã— 5\(\sqrt{2}\)

= 10\(\sqrt{2}\)

Exercise 8.

Simplify \(\sqrt{504}\).

Answer:

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

= \(\sqrt{2^{2}}\) Ã— \(\sqrt{2}\) Ã— \(\sqrt{3^{2}}\) Ã— \(\sqrt{7}\)

= 2 Ã— 3 Ã— \(\sqrt{2}\) Ã— \(\sqrt{7}\)

= 6\(\sqrt{14}\)

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

Simplify each of the square roots in Problems 1â€“5 as much as possible.

Question 1.

\(\sqrt{98}\)

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

= \(\sqrt{2}\) Ã— \(\sqrt{7^{2}}\)

= 7\(\sqrt{2}\)

Question 2.

\(\sqrt{54}\)

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

= \(\sqrt{2}\) Ã— \(\sqrt{3}\) Ã— \(\sqrt{3^{2}}\)

= 3\(\sqrt{6}\)

Question 3.

\(\sqrt{144}\)

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

= 12

Question 4.

\(\sqrt{512}\)

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

= \(\sqrt{2^{2}}\) Ã— \(\sqrt{2^{2}}\) Ã— \(\sqrt{2^{2}}\) Ã— \(\sqrt{2^{2}}\) Ã— \(\sqrt{2}\)

= 2 Ã— 2 Ã— 2 Ã— 2\(\sqrt{2}\)

= 16\(\sqrt{2}\)

Question 5.

\(\sqrt{756}\)

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

= \(\sqrt{2^{2}}\) Ã— \(\sqrt{3^{2}}\) Ã— \(\sqrt{3}\) Ã— \(\sqrt{7}\)

= 2 Ã— 3 Ã— \(\sqrt{21}\)

= 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}

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

\(\sqrt{5^{2}}\) Ã— \(\sqrt{3}\) = 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

\(\sqrt{3^{2}}\) Ã— \(\sqrt{2}\) = 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 = \(\sqrt{2^{2}}\) â‹… \(\sqrt{2^{2}}\) â‹… \(\sqrt{5}\)

x = 2 â‹… 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:

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

= \(\sqrt{2}\) Ã— \(\sqrt{3^{2}}\) Ã— \(\sqrt{5^{2}}\)

= 3 Ã— 5 Ã— \(\sqrt{1}\)

= 15\(\sqrt{1}\)

Yes, Josue is correct because the number 450 = 2 Ã— 3^{2} Ã— 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} Ã— 3^{2} Ã— 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 \(\sqrt{360}\) = \(\sqrt{2^{3}}\) Ã— \(\sqrt{3^{2}}\) Ã— \(\sqrt{5}\). Because we want to simplify square roots, we can rewrite the factor \(\sqrt{2^{3}}\) as \(\sqrt{2^{2}}\) Ã— \(\sqrt{2}\) because of the laws of exponents. Now, we have

\(\sqrt{360}\) = \(\sqrt{2^{2}}\) Ã— âˆš2 Ã— \(\sqrt{3^{2}}\) Ã— \(\sqrt{5}\).

Each perfect square can be simplified as follows:

\(\sqrt{360}\) = 2 Ã— \(\sqrt{2}\) Ã— 3 Ã— \(\sqrt{5}\)

= 2 Ã— 3 Ã— \(\sqrt{2}\) Ã— \(\sqrt{5}\)

= 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:

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

= \(\sqrt{2^{2}}\) Ã— \(\sqrt{6}\)

= 2\(\sqrt{6}\)

Question 2.

\(\sqrt{338}\)

Answer:

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

= \(\sqrt{13^{2}}\) Ã— \(\sqrt{24}\)

= 13\(\sqrt{2}\)

Question 3.

\(\sqrt{196}\)

Answer:

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

= 14

Question 4.

\(\sqrt{2420}\)

Answer:

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

= \(\sqrt{2^{2}}\) Ã— \(\sqrt{11^{2}}\) Ã— \(\sqrt{5}\)

= 2 Ã— 11 Ã— \(\sqrt{5}\)

= 22\(\sqrt{5}\)