Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 10 - Section 10.5 - Rationalizing Numerators and Denominators of Radical Expressions - Exercise Set - Page 717: 21

Answer

$\sqrt{\dfrac{3x}{50}}=\dfrac{\sqrt{6x}}{10}$

Work Step by Step

$\sqrt{\dfrac{3x}{50}}$ Rewrite this expression as $\dfrac{\sqrt{3x}}{\sqrt{25\cdot2}}$ and simplify it: $\sqrt{\dfrac{3x}{50}}=\dfrac{\sqrt{3x}}{\sqrt{25\cdot2}}=\dfrac{\sqrt{3x}}{5\sqrt{2}}=...$ Multiply the fraction by $\dfrac{\sqrt{2}}{\sqrt{2}}$ and simplify again: $...=\dfrac{\sqrt{3x}}{5\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}}=\dfrac{\sqrt{6x}}{5\sqrt{2^{2}}}=\dfrac{\sqrt{6x}}{5(2)}=\dfrac{\sqrt{6x}}{10}$
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.