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: 51

Answer

$-\dfrac{7\sqrt{3}}{12}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To rationalize the given radical expression, $ \dfrac{-7}{\sqrt{48}} ,$ multiply both the numerator and the denominator by an expression that will make the denominator a perfect power of the index. $\bf{\text{Solution Details:}}$ Expressing the radicand with a factor that is a perfect power of the index results to \begin{array}{l}\require{cancel} \dfrac{-7}{\sqrt{16\cdot3}} \\\\ \dfrac{-7}{\sqrt{(4)^2\cdot3}} .\end{array} Multiplying both the numerator and the denominator by an expression that will make the denominator a perfect power of the index results to \begin{array}{l}\require{cancel} \dfrac{-7}{\sqrt{(4)^2\cdot3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}} .\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{-7\sqrt{3}}{\sqrt{(4)^2\cdot3(3)}} \\\\= \dfrac{-7\sqrt{3}}{\sqrt{(4)^2\cdot(3)^2}} \\\\= \dfrac{-7\sqrt{3}}{4\cdot3} \\\\= \dfrac{-7\sqrt{3}}{12} \\\\= -\dfrac{7\sqrt{3}}{12} .\end{array}
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