Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 7 - Section 7.5 - Multiplying and Dividing Radical Expressions - 7.5 Exercises: 62

Answer

$-\dfrac{7r\sqrt{2rs}}{s^3}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To rationalize the given radical expression, $ -\sqrt{\dfrac{98r^3}{s^5}} ,$ multiply the radicand by an expression equal to $1$ which will make the denominator a perfect power of the index. $\bf{\text{Solution Details:}}$ Multiplying the radicand 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{98r^3}{s^5}\cdot\dfrac{s}{s}} \\\\= -\sqrt{\dfrac{98r^3s}{s^6}} .\end{array} Writing the radicand as an expression that contains a factor that is a perfect power of the index results to \begin{array}{l}\require{cancel} -\sqrt{\dfrac{49r^2\cdot2rs}{s^6}} \\\\= -\sqrt{\left(\dfrac{7r}{s^3}\right)^2\cdot2rs} .\end{array} Extracting the root of the factor that is a perfect power of the index results to \begin{array}{l}\require{cancel} -\dfrac{7r}{s^3}\sqrt{2rs} \\\\= -\dfrac{7r\sqrt{2rs}}{s^3} .\end{array}
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