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: 79

Answer

$\dfrac{1}{3+2\sqrt{2}}$

Work Step by Step

Multiplying by the conjugate of the numerator, then the rationalized-numerator form of the given expression, $ \dfrac{\sqrt{2}-1}{\sqrt{2}+1} ,$ is \begin{array}{l}\require{cancel} \dfrac{\sqrt{2}-1}{\sqrt{2}+1}\cdot\dfrac{\sqrt{2}+1}{\sqrt{2}+1} \\\\= \dfrac{(\sqrt{2})^2-1^2}{(\sqrt{2}+1)^2} \\\\= \dfrac{2-1}{(\sqrt{2})^2+2(\sqrt{2})(1)+(1)^2} \\\\= \dfrac{1}{2+2\sqrt{2}+1} \\\\= \dfrac{1}{3+2\sqrt{2}} .\end{array}
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