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: 55

Answer

$\dfrac{12-3\sqrt{2}}{14}$

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{2}+4} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{3}{\sqrt{2}+4}\cdot\dfrac{\sqrt{2}-4}{\sqrt{2}-4} \\\\= \dfrac{3\sqrt{2}-12}{(\sqrt{2})^2-(4)^2} \\\\= \dfrac{3\sqrt{2}-12}{2-16} \\\\= \dfrac{3\sqrt{2}-12}{-14} \\\\= \dfrac{12-3\sqrt{2}}{14} .\end{array}
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