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: 92

Answer

$\sqrt{r}+3$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To rationalize the given radical expression, $ \dfrac{r-9}{\sqrt{r}-3} ,$ 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{r-9}{\sqrt{r}-3} \cdot\dfrac{\sqrt{r}+3}{\sqrt{r}+3} \\\\= \dfrac{(r-9)(\sqrt{r}+3)}{(\sqrt{r}-3)(\sqrt{r}+3)} .\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{(r-9)(\sqrt{r}+3)}{(\sqrt{r})^2-(3)^2} \\\\= \dfrac{(r-9)(\sqrt{r}+3)}{r-9} \\\\= \dfrac{(\cancel{r-9})(\sqrt{r}+3)}{\cancel{r-9}} \\\\= \sqrt{r}+3 .\end{array}
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