Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 10 - Review - Page 748: 114

Answer

$\dfrac{4x^3}{y\sqrt{2x}}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To rationalize the numerator of the given expression, $ \sqrt{\dfrac{24x^5}{3y^2}} ,$ simplify first by extracting the factor that is a perfect power of the index. Then multiply by an expression equal to $1$ which will make the numerator a perfect power of the index. Then extract again the root of the factor that is a perfect power of the index. $\bf{\text{Solution Details:}}$ Simplifying the radicand of the given radical expression results to \begin{array}{l}\require{cancel} \sqrt{\dfrac{8x^5}{y^2}} \\\\= \sqrt{\dfrac{4x^4}{y^2}\cdot2x} \\\\= \sqrt{\left(\dfrac{2x^2}{y}\right)^2\cdot2x} \\\\= \dfrac{2x^2}{y}\sqrt{2x} \\\\= \dfrac{2x^2\sqrt{2x}}{y} .\end{array} Multiplying the given expression by an expression equal to $1$ which will make the numerator a perfect power of the index results to \begin{array}{l}\require{cancel} \dfrac{2x^2\sqrt{2x}}{y}\cdot\dfrac{\sqrt{2x}}{\sqrt{2x}} \\\\= \dfrac{2x^2\sqrt{4x^2}}{y\sqrt{2x}} .\end{array} Extracting the root of the radicand that is a perfect power of the index results to \begin{array}{l}\require{cancel} \dfrac{2x^2\sqrt{(2x)^2}}{y\sqrt{2x}} \\\\= \dfrac{2x^2(2x)}{y\sqrt{2x}} \\\\= \dfrac{4x^3}{y\sqrt{2x}} .\end{array}
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