Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Section 7.5 - Rationalizing Denominators and Numerators of Radical Expressions - Exercise Set: 74

Answer

$\dfrac{7}{\sqrt{15}-1}$

Work Step by Step

Multiplying by the conjugate of the numerator, then the rationalized-numerator form of the given expression, $ \dfrac{\sqrt{15}+1}{2} ,$ is \begin{array}{l}\require{cancel} \dfrac{\sqrt{15}+1}{2}\cdot\dfrac{\sqrt{15}-1}{\sqrt{15}-1} \\\\= \dfrac{(\sqrt{15})^2-(1)^2}{2(\sqrt{15})-2(1)} \\\\= \dfrac{15-1}{2\sqrt{15}-2} \\\\= \dfrac{14}{2\sqrt{15}-2} \\\\= \dfrac{14}{2(\sqrt{15}-1)} \\\\= \dfrac{\cancel{2}\cdot7}{\cancel{2}(\sqrt{15}-1)} \\\\= \dfrac{7}{\sqrt{15}-1} .\end{array}
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