## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter P - Section P.6 - Rational Expressions - Exercise Set - Page 85: 81

#### Answer

The rationalized form of the expression $\frac{\sqrt{x}+\sqrt{y}}{{{x}^{2}}-{{y}^{2}}}$ is $\frac{1}{\left( x+y \right)\left( \sqrt{x}-\sqrt{y} \right)}$.

#### Work Step by Step

Consider the provided expression, $\frac{\sqrt{x}+\sqrt{y}}{{{x}^{2}}-{{y}^{2}}}$ Multiply the numerator and denominator by $\sqrt{x}-\sqrt{y}$ so that the radicand in the numerator is a perfect square. $\frac{\sqrt{x}+\sqrt{y}}{{{x}^{2}}-{{y}^{2}}}=\frac{\sqrt{x}+\sqrt{y}}{{{x}^{2}}-{{y}^{2}}}\cdot \frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}-\sqrt{y}}$ Apply the difference of squares property in the numerator and denominator. Therefore, $\left( \sqrt{x}+\sqrt{y} \right)\left( \sqrt{x}-\sqrt{y} \right)={{\left( \sqrt{x} \right)}^{2}}-{{\left( \sqrt{y} \right)}^{2}}$ And, ${{x}^{2}}-{{y}^{2}}=\left( x-y \right)\left( x+y \right)$ Apply the power of a power property ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ in the expression ${{\left( \sqrt{y} \right)}^{2}},{{\left( \sqrt{x} \right)}^{2}}$. \begin{align} & {{\left( \sqrt{y} \right)}^{2}}={{\left( y \right)}^{2\cdot \frac{1}{2}}} \\ & ={{\left( y \right)}^{1}} \\ & =y \end{align} And, \begin{align} & {{\left( \sqrt{x} \right)}^{2}}={{\left( x \right)}^{2\cdot \frac{1}{2}}} \\ & ={{\left( x \right)}^{1}} \\ & =x \end{align} Rewrite the original expression as: \begin{align} & \frac{\sqrt{x}+\sqrt{y}}{{{x}^{2}}-{{y}^{2}}}=\frac{\sqrt{x}+\sqrt{y}}{{{x}^{2}}-{{y}^{2}}}\cdot \frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}-\sqrt{y}} \\ & =\frac{{{\left( \sqrt{x} \right)}^{2}}-{{\left( \sqrt{y} \right)}^{2}}}{\left( {{x}^{2}}-{{y}^{2}} \right)\left( \sqrt{x}-\sqrt{y} \right)} \\ & =\frac{x-y}{\left( {{x}^{2}}-{{y}^{2}} \right)\left( \sqrt{x}-\sqrt{y} \right)} \\ & =\frac{x-y}{\left( {{x}^{2}}-{{y}^{2}} \right)\left( \sqrt{x}-\sqrt{y} \right)} \end{align} Further simplify, \begin{align} & \frac{\sqrt{x}+\sqrt{y}}{{{x}^{2}}-{{y}^{2}}}=\frac{\left( x-y \right)}{\left( x-y \right)\left( x+y \right)\left( \sqrt{x}-\sqrt{y} \right)} \\ & =\frac{1}{\left( x+y \right)\left( \sqrt{x}-\sqrt{y} \right)} \end{align} Therefore, the rationalized form of the expression $\frac{\sqrt{x}+\sqrt{y}}{{{x}^{2}}-{{y}^{2}}}$ is $\frac{1}{\left( x+y \right)\left( \sqrt{x}-\sqrt{y} \right)}$.

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