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

Published by Pearson

Chapter 10 - Exponents and Radicals - 10.5 Expressions Containing Several Radical Terms - 10.5 Exercise Set - Page 661: 76

Answer

$\dfrac{a-b}{a-2\sqrt{ab}+b}$

Work Step by Step

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

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