## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 7 - Section 7.5 - Multiplying and Dividing Radical Expressions - 7.5 Exercises - Page 475: 57

#### Answer

$\dfrac{2\sqrt{6}}{x}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To rationalize the given radical expression, $\sqrt{\dfrac{24}{x}} ,$ multiply both the numerator and the denominator by an expression that will make the denominator a perfect power of the index. Note that the variables are assumed to represent positive real numbers. $\bf{\text{Solution Details:}}$ 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{24}{x}\cdot\dfrac{x}{x}} \\\\= \sqrt{\dfrac{24x}{x^2}} .\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{4}{x^2}\cdot6} \\\\= \sqrt{\left(\dfrac{2}{x}\right)^2\cdot6} .\end{array} Extracting the root of the factor that is a perfect power of the index results to \begin{array}{l}\require{cancel} \\\\= \dfrac{2}{x}\sqrt{6} \\\\= \dfrac{2\sqrt{6}}{x} .\end{array}

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