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 475: 43

Answer

$\sqrt{7}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To rationalize the given radical expression, $ \dfrac{7}{\sqrt{7}} ,$ multiply both the numerator and the denominator by an expression that will make the denominator a perfect power of the index. $\bf{\text{Solution Details:}}$ 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{7}}\cdot\dfrac{\sqrt{7}}{\sqrt{7}} \\\\= \dfrac{7\sqrt{7}}{\sqrt{7}(\sqrt{7})} .\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{7}}{\sqrt{7(7)}} \\\\= \dfrac{7\sqrt{7}}{\sqrt{7^2}} \\\\= \dfrac{7\sqrt{7}}{7} \\\\= \dfrac{\cancel{7}\sqrt{7}}{\cancel{7}} \\\\= \sqrt{7} .\end{array}
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