Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Section 7.3 - Simplifying Radical Expressions - Exercise Set - Page 433: 56


$2r^{4}s^{6}\sqrt (3r)$

Work Step by Step

$\sqrt (12r^{9}s^{12})=\sqrt (4\times r^{8} \times s^{12}\times 3r)=\sqrt 4\times \sqrt (r^{8})\times \sqrt (s^{12})\times \sqrt (3r)=2r^{4}s^{6}\sqrt (3r)$ We know that $\sqrt 4=2$, because $2^{2}=4$. We also know that $\sqrt (r^{8})=r^{4}$, because $(r^{4})^{2}=r^{4\times2}=r^{8}$ and that $\sqrt (s^{12})=s^{6}$, because $(s^{6})^{2}=s^{6\times2}=s^{12}$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.