Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 10 - Exponents and Radicals - 10.5 Expressions Containing Several Radical Terms - 10.5 Exercise Set: 73

Answer

$\dfrac{2}{3\sqrt{2}+2\sqrt{3}+7\sqrt{6}+14}$

Work Step by Step

Multiplying by the conjugate of the numerator, the rationalized-numerator form of the given expression, $ \dfrac{\sqrt{6}-2}{\sqrt{3}+7} ,$ is \begin{array}{l}\require{cancel} \dfrac{\sqrt{6}-2}{\sqrt{3}+7}\cdot\dfrac{\sqrt{6}+2}{\sqrt{6}+2} \\\\= \dfrac{(\sqrt{6})^2-(2)^2}{\sqrt{3}(\sqrt{6})+\sqrt{3}(2)+7(\sqrt{6})+7(2)} \\\\= \dfrac{6-4}{\sqrt{3(6)}+2\sqrt{3}+7\sqrt{6}+14} \\\\= \dfrac{2}{\sqrt{18}+2\sqrt{3}+7\sqrt{6}+14} \\\\= \dfrac{2}{\sqrt{9\cdot2}+2\sqrt{3}+7\sqrt{6}+14} \\\\= \dfrac{2}{\sqrt{(3)^2\cdot2}+2\sqrt{3}+7\sqrt{6}+14} \\\\= \dfrac{2}{3\sqrt{2}+2\sqrt{3}+7\sqrt{6}+14} .\end{array}
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