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: 94

Answer

$\dfrac{15\sqrt{r}-5\sqrt{s}}{9r-s}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To rationalize the given radical expression, $ \dfrac{5}{3\sqrt{r}+\sqrt{s}} ,$ multiply the numerator and the denominator by the conjugate of the denominator. Then use special products to simplify the result. $\bf{\text{Solution Details:}}$ Multiplying the numerator and the denominator of the given expression by the conjugate of the denominator, the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{5}{3\sqrt{r}+\sqrt{s}}\cdot\dfrac{3\sqrt{r}-\sqrt{s}}{3\sqrt{r}-\sqrt{s}} \\\\= \dfrac{5(3\sqrt{r}-\sqrt{s})}{(3\sqrt{r}+\sqrt{s})(3\sqrt{r}-\sqrt{s})} .\end{array} Using the product of the sum and difference of like terms which is given by $(a+b)(a-b)=a^2-b^2,$ the expression above is equivalent \begin{array}{l}\require{cancel} \dfrac{5(3\sqrt{r}-\sqrt{s})}{(3\sqrt{r})^2-(\sqrt{s})^2} \\\\= \dfrac{5(3\sqrt{r}-\sqrt{s})}{9r-s} .\end{array} Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{5(3\sqrt{r})+5(-\sqrt{s})}{9r-s} \\\\= \dfrac{15\sqrt{r}-5\sqrt{s}}{9r-s} .\end{array}
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