Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 7 - Summary Exercises - Performing Operations with Radicals and Rational Exponents: 11

Answer

$4\sqrt{7}-4\sqrt{5}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To rationalize the given radical expression, $ \dfrac{8}{\sqrt{7}+\sqrt{5}} ,$ 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 by the conjugate of the denominator results to \begin{array}{l}\require{cancel} \dfrac{8}{\sqrt{7}+\sqrt{5}}\cdot\dfrac{\sqrt{7}-\sqrt{5}}{\sqrt{7}-\sqrt{5}} \\\\= \dfrac{8(\sqrt{7}-\sqrt{5})}{(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})} .\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{8(\sqrt{7}-\sqrt{5})}{(\sqrt{7})^2-(\sqrt{5})^2} \\\\= \dfrac{8(\sqrt{7}-\sqrt{5})}{7-5} \\\\= \dfrac{8(\sqrt{7}-\sqrt{5})}{2} \\\\= \dfrac{\cancel{2}(4)(\sqrt{7}-\sqrt{5})}{\cancel{2}} \\\\= 4(\sqrt{7}-\sqrt{5}) .\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} 4(\sqrt{7})+4(-\sqrt{5}) \\\\= 4\sqrt{7}-4\sqrt{5} .\end{array}
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