College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter R - Section R.7 - Radical Expressions - R.7 Exercises: 55

Answer

$\dfrac{\sqrt{6x}}{3x}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the laws of radicals to simplify the given expression, $ \sqrt{\dfrac{2}{3x}} .$ Then rationalize the denominator. $\bf{\text{Solution Details:}}$ Using the Quotient Rule of radicals which is given by $\sqrt[n]{\dfrac{x}{y}}=\dfrac{\sqrt[n]{x}}{\sqrt[n]{y}}{},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{\sqrt{2}}{\sqrt{3x}} .\end{array} Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy}$ to rationalize the denominator, the expression above is equivalent to\begin{array}{l}\require{cancel} \dfrac{\sqrt{2}}{\sqrt{3x}}\cdot\dfrac{\sqrt{3x}}{\sqrt{3x}} \\\\= \dfrac{\sqrt{2(3x)}}{(\sqrt{3x})^2} \\\\= \dfrac{\sqrt{6x}}{3x} .\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.