Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 10 - Exponents and Radicals - 10.4 Dividing Radical Expressions - 10.4 Exercise Set - Page 654: 65

Answer

$\dfrac{2}{\sqrt{6x}}$

Work Step by Step

Rationalizing the numerator, the given expression, $ \dfrac{\sqrt{8}}{2\sqrt{3x}} ,$ is equivalent to \begin{array}{l} \dfrac{\sqrt{8}}{2\sqrt{3x}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}} \\\\= \dfrac{\sqrt{16}}{2\sqrt{6x}} \\\\= \dfrac{4}{2\sqrt{6x}} \\\\= \dfrac{2}{\sqrt{6x}} .\end{array}
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.