Elementary Algebra
17th Edition
ISBN:9780998625713
Author:Lynn Marecek, MaryAnne Anthony-Smith
Publisher:Lynn Marecek, MaryAnne Anthony-Smith
Chapter9: Roots And Radicals
Section9.4: Multiply Square Roots
Problem 312E: A square garden will be made with a stone border on one edge. If only 3+10 feet of stone are...
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Question
![### Fill in the blanks.
#### Rationalize the denominator:
\[ \frac{7}{\sqrt[3]{9a^2}} \]
First, multiply both the numerator and the denominator by \(\sqrt[3]{3a}\):
\[
\frac{7}{\sqrt[3]{9a^2}} = \frac{7 \cdot \sqrt[3]{3a}}{\sqrt[3]{9a^2} \cdot \sqrt[3]{3a}}
\]
The denominator becomes:
\[
\sqrt[3]{9a^2 \cdot 3a} = \sqrt[3]{27a^3}
\]
Since the cube root of \(27a^3\) is simple:
\[
\sqrt[3]{27a^3} = 3a
\]
So the expression simplifies to:
\[
\frac{7 \cdot \sqrt[3]{3a}}{3a}
\]
In summary:
\[
\frac{7}{\sqrt[3]{9a^2}} \cdot \frac{\sqrt[3]{3a}}{\sqrt[3]{3a}} = \frac{7 \sqrt[3]{3a}}{3a}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fef7f77a8-174f-4382-a461-22b0c5210f9d%2Fa37d9df9-9d15-468d-b43e-b0568771191c%2Fd1gy5xc_processed.png&w=3840&q=75)
Transcribed Image Text:### Fill in the blanks.
#### Rationalize the denominator:
\[ \frac{7}{\sqrt[3]{9a^2}} \]
First, multiply both the numerator and the denominator by \(\sqrt[3]{3a}\):
\[
\frac{7}{\sqrt[3]{9a^2}} = \frac{7 \cdot \sqrt[3]{3a}}{\sqrt[3]{9a^2} \cdot \sqrt[3]{3a}}
\]
The denominator becomes:
\[
\sqrt[3]{9a^2 \cdot 3a} = \sqrt[3]{27a^3}
\]
Since the cube root of \(27a^3\) is simple:
\[
\sqrt[3]{27a^3} = 3a
\]
So the expression simplifies to:
\[
\frac{7 \cdot \sqrt[3]{3a}}{3a}
\]
In summary:
\[
\frac{7}{\sqrt[3]{9a^2}} \cdot \frac{\sqrt[3]{3a}}{\sqrt[3]{3a}} = \frac{7 \sqrt[3]{3a}}{3a}
\]
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