Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 8 - Section 8.2 - Multiplying and Dividing Radicals - Exercise Set: 13

Answer

$\sqrt{x}$

Work Step by Step

Using $\sqrt{x}\cdot\sqrt{y}=\sqrt{xy}$ or the product rule of radicals, the given expression, $ \sqrt{\dfrac{2x}{9}}\cdot\sqrt{\dfrac{9}{2}} ,$ is equivalent to \begin{array}{l}\require{cancel} \sqrt{\dfrac{2x}{9}\cdot\dfrac{9}{2}} \\\\= \sqrt{\dfrac{\cancel{2}x}{\cancel{9}}\cdot\dfrac{\cancel{9}}{\cancel{2}}} \\\\= \sqrt{x} .\end{array}
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