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: 57

Answer

$4\sqrt{6}+8$

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{8}{\sqrt{6}-2} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{8}{\sqrt{6}-2}\cdot\dfrac{\sqrt{6}+2}{\sqrt{6}+2} \\\\= \dfrac{8(\sqrt{6}+2)}{(\sqrt{6})^2-(2)^2} \\\\= \dfrac{8(\sqrt{6}+2)}{6-4} \\\\= \dfrac{8(\sqrt{6}+2)}{2} \\\\= \dfrac{\cancel{2}(4)(\sqrt{6}+2)}{\cancel{2}} \\\\= 4(\sqrt{6}+2) \\\\= 4\sqrt{6}+8 .\end{array}
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