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: 24

Answer

$\dfrac{1}{\sqrt{32x}}=\dfrac{\sqrt{2x}}{8x}$

Work Step by Step

$\dfrac{1}{\sqrt{32x}}$ Multiply the fraction by $\dfrac{\sqrt{32x}}{\sqrt{32x}}$: $\dfrac{1}{\sqrt{32x}}=\dfrac{1}{\sqrt{32x}}\cdot\dfrac{\sqrt{32x}}{\sqrt{32x}}=\dfrac{\sqrt{32x}}{\sqrt{(32x)^{2}}}=\dfrac{\sqrt{32x}}{32x}=...$ Rewrite the expression as $\dfrac{\sqrt{16\cdot2\cdot x}}{32x}$ and simplify: $...=\dfrac{\sqrt{16\cdot2\cdot x}}{32x}=\dfrac{4\sqrt{2x}}{32x}=\dfrac{\sqrt{2x}}{8x}$
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.