# Eureka Math Grade 8 Module 7 Lesson 4 Answer Key

## 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.
$$\sqrt{50}$$ =
→ 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 × 52
Since 50 = 2 × 52, 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, 52 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 × 52 ) = $$\sqrt{50}$$.

Example 2.
Simplify the square root as much as possible.
$$\sqrt{28}$$ =
→ 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 = 22 × 7
Since 28 = 22 × 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, 22 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.
$$\sqrt{128}$$ =
→ 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 = 27
→ Since 128 = 27, then $$\sqrt{128}$$ = √(27 ). We know that we can simplify perfect squares, so we can rewrite 27 as 22 × 22 × 22 × 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 22 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.
$$\sqrt{288}$$ =
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 = 25 × 32
Since 288 = 25 × 32, then $$\sqrt{288}$$ = √(25 × 32 ). What do we do next?
Use the laws of exponents to rewrite 25 as 22 × 22 × 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?
4

ii. What does 4 × 4 equal?
16

iii. Does $$\sqrt{16}$$ = $$\sqrt{4 \times 4}$$?
Yes

b.
i. What does $$\sqrt{36}$$ equal?
6

ii. What does 6 × 6 equal?
36

iii. Does $$\sqrt{36}$$ = $$\sqrt{6 \times 6}$$?
Yes

c.
i. What does $$\sqrt{121}$$ equal?
11

ii. What does 11 × 11 equal?
121

iii. Does $$\sqrt{121}$$ = $$\sqrt{11 \times 11}$$?
Yes

d.
i. What does $$\sqrt{81}$$ equal?
9

ii. What does 9 × 9 equal?
81

iii. Does $$\sqrt{81}$$ = $$\sqrt{9 \times 9}$$?
Yes

e. Rewrite $$\sqrt{20}$$ using at least one perfect square factor.
$$\sqrt{20}$$ = $$\sqrt{4 \times 5}$$

f. Rewrite $$\sqrt{28}$$ using at least one perfect square factor.
$$\sqrt{28}$$ = $$\sqrt{4 \times 7}$$

Exercises 1–4
Simplify the square roots as much as possible.

Exercise 1.
$$\sqrt{18}$$
$$\sqrt{18}$$ = $$\sqrt{2 \times 3^{2}}$$
= $$\sqrt{2}$$ × $$\sqrt{3^{2}}$$
= 3$$\sqrt{2}$$

Exercise 2.
$$\sqrt{44}$$
$$\sqrt{44}$$ = $$\sqrt{2^{2} \times 11}$$
= $$\sqrt{2^{2}}$$ × $$\sqrt{11}$$
= 2$$\sqrt{11}$$

Exercise 3.
$$\sqrt{169}$$
$$\sqrt{169}$$ = $$\sqrt{13^{2}}$$
= 13

Exercise 4.
$$\sqrt{75}$$
$$\sqrt{75}$$ = $$\sqrt{3 \times 5^{2}}$$
= $$\sqrt{3}$$ × $$\sqrt{5^{2}}$$
= 5$$\sqrt{3}$$

Exercises 5–8

Exercise 5.
Simplify $$\sqrt{108}$$.
$$\sqrt{108}$$ = $$\sqrt{2^{2} \times 3^{3}}$$
= $$\sqrt{2^{2}}$$ × $$\sqrt{3^{2}}$$ × $$\sqrt{3}$$
= 2 × 3$$\sqrt{3}$$
= 6$$\sqrt{3}$$

Exercise 6.
Simplify $$\sqrt{250}$$.
$$\sqrt{250}$$ = $$\sqrt{2 \times 5^{3}}$$
= $$\sqrt{2}$$ × $$\sqrt{5^{2}}$$ × $$\sqrt{5}$$
= 5$$\sqrt{2}$$ × $$\sqrt{5}$$
= 5$$\sqrt{10}$$

Exercise 7.
Simplify $$\sqrt{200}$$.
$$\sqrt{200}$$ = $$\sqrt{2^{3} \times 5^{2}}$$
= $$\sqrt{2^{2}}$$ × $$\sqrt{2}$$ × $$\sqrt{5^{2}}$$
= 2 × 5$$\sqrt{2}$$
= 10$$\sqrt{2}$$

Exercise 8.
Simplify $$\sqrt{504}$$.
$$\sqrt{504}$$ = $$\sqrt{2^{3} \times 3^{2} \times 7}$$
= $$\sqrt{2^{2}}$$ × $$\sqrt{2}$$ × $$\sqrt{3^{2}}$$ × $$\sqrt{7}$$
= 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. Let c units represent the length of the hypotenuse.
($$\sqrt{27}$$)2 + ($$\sqrt{48}$$)2 = c2
27 + 48 = c2
75 = c2
$$\sqrt{75}$$ = $$\sqrt{c^{2}}$$
$$\sqrt{5^{2}}$$ × $$\sqrt{3}$$ = 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. Let c cm represent the length of the hypotenuse.
32 + 82 = c2
9 + 64 = c2
73 = c2
$$\sqrt{73}$$ = $$\sqrt{c^{2}}$$
$$\sqrt{73}$$ = 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. Let c mm represent the length of the hypotenuse.
32 + 32 = c2
9 + 9 = c2
18 = c2
$$\sqrt{18}$$ = $$\sqrt{c^{2}}$$
$$\sqrt{18}$$ = 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. Let x in. represent the unknown length.
x2 + 82 = 122
x2 + 64 = 144
x2 + 64 – 64 = 144 – 64
x2 = 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.
$$\sqrt{450}$$ = $$\sqrt{2 \times 3^{2} \times 5^{2}}$$
= $$\sqrt{2}$$ × $$\sqrt{3^{2}}$$ × $$\sqrt{5^{2}}$$
= 3 × 5 × $$\sqrt{1}$$
= 15$$\sqrt{1}$$
Yes, Josue is correct because the number 450 = 2 × 32 × 52. 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.
To simplify $$\sqrt{360}$$, first write the factors of 360. The number 360 = 23 × 32 × 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.
$$\sqrt{24}$$
$$\sqrt{24}$$ = $$\sqrt{2^{2} \times 6}$$
= $$\sqrt{2^{2}}$$ × $$\sqrt{6}$$
= 2$$\sqrt{6}$$

Question 2.
$$\sqrt{338}$$
$$\sqrt{338}$$ = $$\sqrt{13^{2} \times 2}$$
= $$\sqrt{13^{2}}$$ × $$\sqrt{24}$$
= 13$$\sqrt{2}$$

Question 3.
$$\sqrt{196}$$
$$\sqrt{196}$$ = $$\sqrt{14^{2}}$$
$$\sqrt{2420}$$
$$\sqrt{2420}$$ = $$\sqrt{2^{2} \times 11^{2} \times 5}$$
= $$\sqrt{2^{2}}$$ × $$\sqrt{11^{2}}$$ × $$\sqrt{5}$$
= 2 × 11 × $$\sqrt{5}$$
= 22$$\sqrt{5}$$