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 - Page 476: 66

Answer

$\dfrac{2\sqrt{5rm}}{m^2}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To rationalize the given radical expression, $ \dfrac{2\sqrt{5r}}{\sqrt{m^3}} ,$ multiply the numerator and the denominator by an expression equal to $1$ which will make the denominator a perfect power of the index. $\bf{\text{Solution Details:}}$ Multiplying the numerator and the denominator by an expression equal to $1$ which will make the denominator a perfect power of the index results to \begin{array}{l}\require{cancel} \dfrac{2\sqrt{5r}}{\sqrt{m^3}}\cdot\dfrac{\sqrt{m}}{\sqrt{m}} .\end{array} Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the expression above is equivalent to\begin{array}{l}\require{cancel} \dfrac{2\sqrt{5rm}}{\sqrt{m^3(m)}} \\\\= \dfrac{2\sqrt{5rm}}{\sqrt{m^4}} \\\\= \dfrac{2\sqrt{5rm}}{\sqrt{(m^2)^2}} .\end{array} Extracting the root of the radicand that is a perfect power of the index results to \begin{array}{l}\require{cancel} \dfrac{2\sqrt{5rm}}{m^2} .\end{array}
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