## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 7 - Section 7.5 - Multiplying and Dividing Radical Expressions - 7.5 Exercises - Page 475: 59

#### Answer

$\dfrac{-8\sqrt{3k}}{k}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To rationalize the given radical expression, $\dfrac{-8\sqrt{3}}{\sqrt{k}} ,$ multiply both the numerator and the denominator by the expression equal to the denominator. $\bf{\text{Solution Details:}}$ Multiplying both the numerator and the denominator by the expression equal to the denominator results to \begin{array}{l}\require{cancel} \dfrac{-8\sqrt{3}}{\sqrt{k}}\cdot\dfrac{\sqrt{k}}{\sqrt{k}} \\\\= \dfrac{-8\sqrt{3}(\sqrt{k})}{(\sqrt{k})^2} \\\\= \dfrac{-8\sqrt{3}(\sqrt{k})}{k} .\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{-8\sqrt{3(k)}}{k} \\\\= \dfrac{-8\sqrt{3k}}{k} .\end{array}

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