Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 9 - Roots and Radicals - 9.4 - Products and Quotients Involving Radicals - Problem Set 9.4 - Page 418: 60



Work Step by Step

Multiplying by the conjugate of the denominator and using $(a+b)(a-b)=a^2-b^2,$ the given expression, $ \dfrac{3}{\sqrt{6}+\sqrt{5}} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{3}{\sqrt{6}+\sqrt{5}}\cdot\dfrac{\sqrt{6}-\sqrt{5}}{\sqrt{6}-\sqrt{5}} \\\\= \dfrac{3\sqrt{6}-3\sqrt{5}}{(\sqrt{6})^2-(\sqrt{5})^2} \\\\= \dfrac{3\sqrt{6}-3\sqrt{5}}{6-5} \\\\= \dfrac{3\sqrt{6}-3\sqrt{5}}{1} \\\\= 3\sqrt{6}-3\sqrt{5} .\end{array}
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