Engage NY Eureka Math Algebra 2 Module 3 Lesson 3 Answer Key

Eureka Math Algebra 2 Module 3 Lesson 3 Opening Exercise Answer Key

Opening Exercise:

a. What is the value of 2\(\frac{1}{2}\)? Justify your answer.
Answer:
A possible student response follows: I think it will be around 1.5 because 20 = 1 and 21 = 2.

b. Graph f(x) = 2x for each integer x from x = -2 to x = 5. Connect the points on your graph with a smooth curve.

Eureka Math Algebra 2 Module 3 Lesson 3 Opening Exercise Answer Key 1

Answer:

Eureka Math Algebra 2 Module 3 Lesson 3 Opening Exercise Answer Key 2

c. The graph on the right shows a close-up view of f(x) 2x for -0. 5 < x < 1.5.

Eureka Math Algebra 2 Module 3 Lesson 3 Opening Exercise Answer Key 3

c. Find two consecutive integers that are under and over estimates of the value of 2\(\frac{1}{2}\).
Answer:
20 < 2\(\frac{1}{2}\) < 21

1 < 2\(\frac{1}{2}\) < 2

d. Does it appear that 2\(\frac{1}{2}\) is halfway between the integers you specified in Exercise 1?
Answer:
No. It looks like 2\(\frac{1}{2}\) is a little less than halfway between 1 and 2.

e. Use the graph of f(x) = 2 to estimate the value of 2\(\frac{1}{2}\).
Answer:
2\(\frac{1}{2}\) ≈ 1.4

f. Use the graph of f(x) = 2x to estimate the value of 2\(\frac{1}{3}\).
Answer:
2\(\frac{1}{3}\) ≈ 1.25

Eureka Math Algebra 2 Module 3 Lesson 3 Example Answer Key

a. What is the 4th root of 16?
Answer:
x4 = 16 when x = 2 because 24 = 16. Thus, \(\sqrt[4]{16}\) = 2.

b. What is the cube root of 125?
Answer:
x3 = 125 when x = 5 because 53 = 125. Thus, \(\sqrt[3]{125}\) = 5.

c. What is the 5th root of 100,000?
Answer:
x5 = 100,000 when x = 10 because 105 = 100000. Thus, \(\sqrt[5]{100,000}\) = 10.

Eureka Math Algebra 2 Module 3 Lesson 3 Exercise Answer Key

Exercise 1:

Exercise 1.
Evaluate each expression.

a. \(\sqrt[4]{16}\)
Answer:
3

b. \(\sqrt[5]{32}\)
Answer:
2

c. \(\sqrt[3]{9} \cdot \sqrt[3]{3}\)
Answer:
\(\sqrt[3]{27}\) = 3

d. \(\sqrt[4]{25} \cdot \sqrt[4]{100} \cdot \sqrt[4]{4}\)
Answer:
\(\sqrt[4]{10,000}\) = 10

Discussion:

1.
If 2\(\frac{1}{2}\) = √2 and 2\(\frac{1}{3}\) = \(\sqrt[3]{2}\) what does 2\(\frac{3}{4}\) equal? Explain your reasoning.
Answer:
Student solutions and explanations will vary. One possible solution would be 2\(\frac{3}{4}\) = (2\(\frac{1}{4}\))3 , soit must mean that 2\(\frac{3}{4}\) = \((\sqrt[4]{2})^{3}\). Since the properties of exponents and the meaning of an exponent made sense with integers and now for rational numbers in the form \(\frac{1}{n}\), it would make sense that they would work for all rational numbers, too.

Exercises 2 – 12:

Rewrite each exponential expression as a radical expression.

Exercise 2.
3\(\frac{1}{2}\)
Answer:
3\(\frac{1}{2}\) = √3

Exercise 3.
11\(\frac{1}{5}\)
Answer:
11\(\frac{1}{5}\) = \(\sqrt[5]{11}\)

Exercise 4.
\(\left(\frac{1}{4}\right)^{\frac{1}{5}}\)
Answer:
\(\left(\frac{1}{4}\right)^{\frac{1}{5}}\) = \(\sqrt[5]{\frac{1}{4}}\)

Exercise 5.
6\(\frac{1}{10}\)
Answer:
6\(\frac{1}{10}\) = \(\sqrt[10]{6}\)

Rewrite the following exponential expressions as equivalent radical expressions. If the number is rational, write it without radicals or exponents.

Exercise 6.
2\(\frac{3}{2}\)
Answer:
2\(\frac{3}{2}\) = \(\sqrt{2^{3}}\) = 2√2

Exercise 7.
4\(\frac{5}{2}\)
Answer:
4\(\frac{5}{2}\) = \(\sqrt{4^{5}}\) = (√4)5 = 25 = 32

Exercise 8.
\(\left(\frac{1}{8}\right)^{\frac{5}{3}}\)
Answer:
\(\left(\frac{1}{8}\right)^{\frac{5}{3}}\) = \(\sqrt[3]{\left(\frac{1}{8}\right)^{5}}\)
= \(\left(\sqrt[3]{\frac{1}{8}}\right)^{5}\) = (\(\frac{1}{2}\))5
= \(\frac{1}{32}\)

Exercise 9.
Show why the following statement is true:
\(2^{-\frac{1}{2}}=\frac{1}{2^{\frac{1}{2}}}\)
Answer:
Student solutions and explanations will vary. One possible solution would be

Eureka Math Algebra 2 Module 3 Lesson 3 Exercise Answer Key 4

Rewrite the following exponential expressions as equivalent radical expressions. If the number is rational, write it without radicals or exponents.

Exercise 10.
4\(-\frac{3}{2}\)
Answer:
4\(-\frac{3}{2}\) = \(\frac{1}{\sqrt{4^{3}}}=\frac{1}{(\sqrt{4})^{3}}=\frac{1}{8}\)

Exercise 11.
27\(-\frac{2}{3}\)
Answer:
27\(-\frac{2}{3}\) = \(=\frac{1}{27^{\frac{2}{3}}}=\frac{1}{(\sqrt[3]{27})^{2}}=\frac{1}{3^{2}}=\frac{1}{9}\)

Exercise 12.
\(\left(\frac{1}{4}\right)^{-\frac{1}{2}}\)
Answer:
we have \(\left(\frac{1}{4}\right)^{-\frac{1}{2}}\) = \(\left(\sqrt{\frac{1}{4}}\right)^{-1}\) = \(\left(\frac{1}{2}\right)^{-1}\) = 2,

Alternatively, \(\left(\frac{1}{4}\right)^{-\frac{1}{2}}\) = \(\left(\left(\frac{1}{4}\right)^{-1}\right)^{\frac{1}{2}}\)
= (4)\(\frac{1}{2}\) = √4 = 2.

Eureka Math Algebra 2 Module 3 Lesson 3 Problem Set Answer Key

Question 1.
Select the expression from (A), (B), and (C) that correctly completes the statement.

a. x\(\frac{1}{3}\) is equivalent to ____.
(A) \(\frac{1}{3}\)x
(B) \(\sqrt[3]{x}\)
(C) \(\frac{3}{x}\)
Answer:
(B)

b. x\(\frac{2}{3}\)
(A) \(\frac{2}{3}\)x
(B) \(\sqrt[3]{x^{2}}\)
(C) \((\sqrt{x})^{3}\)
Answer:
(B)

c \(x^{-\frac{1}{4}}\)
(A) –\(\frac{1}{4}\)x
(B) \(\frac{4}{x}\)
(C) \(\frac{1}{\sqrt[4]{x}}\)
Answer:
(C)

d. \(\left(\frac{4}{x}\right)^{\frac{1}{2}}\)
(A) \(\frac{2}{x}\)
(B) \(\frac{4}{x^{2}}\)
(C) \(\frac{2}{\sqrt{x}}\)
Answer:
(C)

Question 2.
Identify which of the expressions (A), (B), and (C) are equivalent to the given expression.

a. \(16^{\frac{1}{2}}\)
(A) \(\left(\frac{1}{16}\right)^{-\frac{1}{2}}\)
(B) \(8^{\frac{2}{3}}\)
(C) \(64^{\frac{3}{2}}\)
Answer:
(A) and (B)

b. \(\left(\frac{2}{3}\right)^{-1}\)
(A) \(-\frac{3}{2}\)
(B) \(\left(\frac{9}{4}\right)^{\frac{1}{2}}\)
(C) \(\frac{27^{\frac{1}{3}}}{6}\)
Answer:
(B) only

Question 3.
Rewrite in radical form. If the number is rational, write it without using radicals.

a. \(6^{\frac{3}{2}}\)
Answer:
√216

b. \(\left(\frac{1}{2}\right)^{\frac{1}{4}}\)
Answer:
\(\sqrt[4]{\frac{1}{2}}\)

c. 3\((8)^{\frac{1}{3}}\)
Answer:
\(3 \sqrt[3]{8}\) = 6

d. \(\left(\frac{64}{125}\right)^{-\frac{2}{3}}\)
Answer:
\(\left(\sqrt[3]{\frac{125}{64}}\right)^{2}=\frac{25}{16}\)

e. \(81^{-\frac{1}{4}}\)
Answer:
\(\frac{1}{\sqrt[4]{81}}=\frac{1}{3}\)

Question 4.
Rewrite the following expressions in exponent form.

a. √5
Answer:
5\(\frac{1}{2}\)

b. \(\sqrt[3]{5^{2}}\)
Answer:
5\(\frac{2}{3}\)

c. \(\sqrt{5^{3}}\)
Answer:
5\(\frac{3}{2}\)

d. \((\sqrt[3]{5})^{2}\)
Answer:
5\(\frac{2}{3}\)

Question 5.
Use the graph of f(x) = 2X shown to the right to estimate the following powers of 2.

Eureka Math Algebra 2 Module 3 Lesson 3 Problem Set Answer Key 5

a. 2\(\frac{1}{4}\)
Answer:
≈ 1.2

b. 2\(\frac{2}{3}\)
Answer:
≈ 1.6

c. 2\(\frac{3}{4}\)
Answer:
≈ 1.7

d. 20.2
Answer:
≈ 1.15

e. 21.2
Answer:
≈ 2.3

f. 2–\(\frac{1}{5}\)
Answer:
≈ 0.85

Question 6.
Rewrite each expression in the form kxn, where k is a real number, x is a positive real number, and n is rational.

a. \(\sqrt[4]{16 x^{3}}\)
Answer:
\(2 x^{\frac{3}{4}}\)

b.\(\frac{5}{\sqrt{x}}\)
Answer:
\(5 x^{-\frac{1}{2}}\)

c. \(\sqrt[3]{ x}\)
Answer:
\(x^{-\frac{4}{3}}\)

d.\(\frac{4}{\sqrt[3]{8 x^{3}}}\)
Answer:
\(2 x^{-1}\)

e. \(\frac{27}{\sqrt{9 x^{4}}}\)
Answer:
\(9 x^{-2}\)

f. \(\left(\frac{125}{x^{2}}\right)^{-\frac{1}{3}}\)
Answer:
\(\frac{1}{5} x^{\frac{2}{3}}\)

Question 7.
Find the value of x for which 2\(x^{\frac{1}{2}}\) = 32
Answer:
256

Question 8.
Find the value of x for which x\(x^{\frac{4}{3}}\) = 81
Answer:
27

Question 9.
If \(x^{\frac{3}{2}}\) = 64, find the value of 4\(x^{\frac{3}{4}}\).
Answer:
x = 16, so 4\((16)^{-\frac{3}{4}}\) = 4(2)-3 = \(\frac{4}{8}\) = \(\frac{1}{2}\)

Question 10.
Evaluate the following expressions when b = \(\frac{1}{9}\)

a. \(b^{-\frac{1}{2}}\)
Answer:
\(\left(\frac{1}{9}\right)^{-\frac{1}{2}} = 9^{\frac{1}{2}}\) = 3

b. \(b^{\frac{5}{2}}\)
Answer:
\(\left(\frac{1}{9}\right)^{\frac{5}{2}}=\left(\frac{1}{3}\right)^{5}=\frac{1}{243}\)

c. \(\sqrt[3]{3 b^{-1}}\)
Answer:
\(\sqrt[3]{3\left(\frac{1}{9}\right)^{-1}}=\sqrt[3]{27}\) = 3

Question 11.
Show that each expression is equivalent to 2x. Assume x is a positive real number.

a. \(\sqrt[4]{16 x^{4}}\)
Answer:
\(\sqrt[4]{16} \cdot \sqrt[4]{x^{4}}\) = 2x

b. \(\frac{\left(\sqrt[3]{8 x^{3}}\right)^{2}}{\sqrt{4 x^{2}}}\)
Answer:
\(\frac{(2 x)^{2}}{\left(4 x^{2}\right)^{\frac{1}{2}}}=\frac{4 x^{2}}{2 x}\) = 2x

c. \(\frac{6 x^{3}}{\sqrt[3]{27 x^{6}}}\)
Answer:
\(\frac{6 x^{3}}{3 x^{\frac{6}{3}}}=\frac{6 x^{3}}{3 x^{2}}\) = 2x

Question 12.
Yoshiko said that 164 = 4 because 4 is one-fourth of 16. Use properties of exponents to explain why she is or is not correct.
Answer:
Yoshiko’s reasoning is not correct. By our exponent properties, \(\left(16^{\frac{1}{4}}\right)^{4}\) = \(16^{\left(\frac{1}{4}\right) \cdot 4}\) = 161 = 16, but 44 = 256.
Since \(\left(16^{\frac{1}{4}}\right)^{4}\) ≠ 44, we know that \(16^{\frac{1}{4}}\) ≠ 4.

Question 13.
Jefferson said that \(8^{\frac{4}{3}}\) = 16 because \(8^{\frac{1}{3}}\) = 2 and 24 = 16. Use properties of exponents to explain why he is or is not correct.
Answer:
Jefferson’s reasoning is correct. We know that \(8^{\frac{4}{3}}\) = \(\left(8^{\frac{1}{3}}\right)^{4}\), so \(8^{\frac{4}{3}}\) = 24, and thus \(8^{\frac{4}{3}}\) = 16.

Question 14.
Rita said that \(8^{\frac{2}{3}}\) = 128 because \(8^{\frac{2}{3}}\) = 82 \(8^{\frac{1}{3}}\), so \(8^{\frac{2}{3}}\) = 64 . 2, and then \(8^{\frac{2}{3}}\) = 128. Use properties of exponents to explain why she is or is not correct.
Answer:
Rita’s reasoning is not correct because she did not apply the properties of exponents correctly. She should also realize that raising 8 to a positive power less than 1 would produce a number less than 8. The correct calculation is below.

\(8^{\frac{2}{3}}\) = \(\left(8^{\frac{1}{3}}\right)^{2}\) = 22 = 4

Question 15.
Suppose for some positive real number a that \(\left(a^{\frac{1}{4}} \cdot a^{\frac{1}{2}} \cdot a^{\frac{1}{4}}\right)^{2}\) = 3.

a. What is the value of a?
Answer:

Eureka Math Algebra 2 Module 3 Lesson 3 Problem Set Answer Key 6

b. Which exponent properties did you use to find your answer to part (a)?
Answer:
We used the properties bn . bm = bm + n and b\(\left(\boldsymbol{b}^{m}\right)^{n}\) = bmn

Question 16.
In the lesson, you made the following argument:
Eureka Math Algebra 2 Module 3 Lesson 3 Problem Set Answer Key 7

Since \(\sqrt[3]{2}\) is a number so that (\(\sqrt[3]{2}\))3 = 2 and \(2^{\frac{1}{3}}\) is a number so that \(\left(2^{\frac{1}{3}}\right)^{3}\) = 2, you concluded that \(2^{\frac{1}{3}}\) = \(\sqrt[3]{2}\). Which exponent property was used to make this argument?
Answer:
We used the property bn . bm = bm + n. (Students may also mention the uniqueness of nth roots.)

Eureka Math Algebra 2 Module 3 Lesson 3 Exit Ticket Answer Key

Question 1.
Rewrite the following exponential expressions as equivalent radical expressions.

a. \(2^{\frac{1}{2}}\)
Answer:
\(2^{\frac{1}{2}}\) = √2

b. \(2^{\frac{3}{4}}\)
Answer:
\(2^{\frac{3}{4}}\) = \(\sqrt[4]{2^{3}}\) = \(\sqrt[4]{8}\)

c. \(3^{-\frac{2}{3}}\)
Answer:
\(3^{-\frac{2}{3}}\) = \(\frac{1}{\sqrt[3]{3^{2}}}=\frac{1}{\sqrt[3]{9}}\)

Question 2.
Rewrite the following radical expressions as equivalent exponential expressions.

a. √5
Answer:
√5 = \(5^{\frac{1}{2}}\)

b. \(2 \sqrt[4]{3}\)
Answer:
\(2 \sqrt[4]{3}\) = \(\sqrt[4]{2^{4} \cdot 3}=\sqrt[4]{48}=48^{\frac{1}{4}}\)

c. \(\frac{1}{\sqrt[3]{16}}\)
Answer:
\(\frac{1}{\sqrt[3]{16}}=\left(2^{4}\right)^{-\frac{1}{3}}=2^{-\frac{4}{3}}\)
\(\frac{1}{\sqrt[3]{16}}=(16)^{-\frac{1}{3}}\)

Question 3.
Provide a written explanation for each question below.

a. Is it true that \(\left(1000^{\frac{1}{3}}\right)^{3}=\left(1000^{3}\right)^{\frac{1}{3}}\)? Explain how you know.
Answer:
\(\left(1000^{\frac{1}{3}}\right)^{3}=(\sqrt[3]{1000})^{3}\)
= 103 = 1000
\(\left(1000^{3}\right)^{\frac{1}{3}}=(1000000000)^{\frac{1}{3}}\) = 1000
So, this statement is true.

b. Is it true that \(\left(4^{\frac{1}{2}}\right)^{3}=\left(4^{3}\right)^{\frac{1}{2}}\)? Explain how you know.
Answer:
\(\left(4^{\frac{1}{2}}\right)^{3}\) = (√4)3 = 23 = 8
\(\left(4^{3}\right)^{\frac{1}{2}}=64^{\frac{1}{2}}\) = √64 = 8
So, this statement is true.

c. Suppose that m and n are positive integers and b is a real number so that the principal nth root of b exists. In general, does \(\left(b^{\frac{1}{n}}\right)^{m}=\left(b^{m}\right)^{\frac{1}{n}}\)? Explain how you know.
Answer:
From the two examples we have seen, it appears that we can extend the property (\(\left(b^{\frac{1}{n}}\right)^{m}=\left(b^{m}\right)^{\frac{1}{n}}\) for integers m and n to rational exponents.

We know that, in general, we have
\(\left(b^{\frac{1}{n}}\right)^{m}=(\sqrt[n]{b})^{m}\)
= \(\sqrt[n]{\boldsymbol{b}^{m}}\)
= \(\left(\boldsymbol{b}^{m}\right)^{\frac{1}{n}}\)

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