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 - Review Exercises: Chapter 10: 40

Answer

$\dfrac{\sqrt{2xy}}{4y}$

Work Step by Step

Multiplying by an expression equal to $1$ which will make the denominator a perfect power of the index results to \begin{array}{l}\require{cancel} \sqrt{\dfrac{x}{8y}} \\\\= \sqrt{\dfrac{x}{8y}\cdot\dfrac{2y}{2y}} \\\\= \sqrt{\dfrac{1}{16y^2}\cdot2xy} \\\\= \sqrt{ \left( \dfrac{1}{4y} \right)^2\cdot2xy} \\\\= \dfrac{1}{4y}\sqrt{2xy} \\\\= \dfrac{\sqrt{2xy}}{4y} .\end{array}
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