## Intermediate Algebra (6th Edition)

Published by Pearson

# Chapter 7 - Section 7.5 - Rationalizing Denominators and Numerators of Radical Expressions - Exercise Set: 47

#### Answer

$\dfrac{a+\sqrt{ab}+\sqrt{a}+\sqrt{b}}{4a-b}$

#### Work Step by Step

Rationalizing the denominator of $\dfrac{\sqrt{a}+1}{2\sqrt{a}-\sqrt{b}}$ results to \begin{array}{l} \dfrac{\sqrt{a}+1}{2\sqrt{a}-\sqrt{b}} \cdot \dfrac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}+\sqrt{b}} \\\\= \dfrac{(\sqrt{a})(\sqrt{a})+(\sqrt{a})(\sqrt{b})+(1)(\sqrt{a})+(1)(\sqrt{b})}{(2\sqrt{a})^2-(\sqrt{b})^2} \\\\= \dfrac{(\sqrt{a})(\sqrt{a})+(\sqrt{a})(\sqrt{b})+(1)(\sqrt{a})+(1)(\sqrt{b})}{(2\sqrt{a})^2-(\sqrt{b})^2} \\\\= \dfrac{a+\sqrt{ab}+\sqrt{a}+\sqrt{b}}{4a-b} .\end{array}

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