## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 7 - Section 7.5 - Multiplying and Dividing Radical Expressions - 7.5 Exercises: 88

#### Answer

$\dfrac{3\sqrt{2}+2\sqrt{7}}{10}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To rationalize the given radical expression, $\dfrac{-1}{3\sqrt{2}-2\sqrt{7}} ,$ 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{-1}{3\sqrt{2}-2\sqrt{7}} \cdot\dfrac{3\sqrt{2}+2\sqrt{7}}{3\sqrt{2}+2\sqrt{7}} \\\\= \dfrac{-1(3\sqrt{2}+2\sqrt{7})}{(3\sqrt{2}-2\sqrt{7})(3\sqrt{2}+2\sqrt{7})} .\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{-1(3\sqrt{2}+2\sqrt{7})}{(3\sqrt{2})^2-(2\sqrt{7})^2} \\\\= \dfrac{-1(3\sqrt{2}+2\sqrt{7})}{9(2)-4(7)} \\\\= \dfrac{-1(3\sqrt{2}+2\sqrt{7})}{18-28} \\\\= \dfrac{-1(3\sqrt{2}+2\sqrt{7})}{-10} \\\\= \dfrac{1(3\sqrt{2}+2\sqrt{7})}{10} .\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{1(3\sqrt{2})+1(2\sqrt{7})}{10} \\\\= \dfrac{3\sqrt{2}+2\sqrt{7}}{10} .\end{array}

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