## Intermediate Algebra for College Students (7th Edition)

Published by Pearson

# Chapter 7 - Section 7.5 - Multiplying with More Than One Term and Rationalizing Denominators - Exercise Set - Page 550: 83

#### Answer

$25\sqrt{2}+15\sqrt{5}$

#### Work Step by Step

We lose the square roots in the denominator by applying the difference of squares formula: $(a\sqrt{x}+b\sqrt{y})(a\sqrt{x}-b\sqrt{y})=(a\sqrt{x})^{2}-(b\sqrt{y})^{2}=a^{2}x-b^{2}y$ $\displaystyle \frac{25}{5\sqrt{2}-3\sqrt{5}} \displaystyle \color{red}{ \cdot\frac{5\sqrt{2}+3\sqrt{5}}{5\sqrt{2}+3\sqrt{5}} }\qquad$ (rationalize) $=\displaystyle \frac{25(5\sqrt{2}+3\sqrt{5})}{(5\sqrt{2})^{2}-(3\sqrt{5})^{2}} =\frac{25(5\sqrt{2}+3\sqrt{5})}{5^{2}\cdot 2-3^{2}\cdot 5}$ $=\displaystyle \frac{25(5\sqrt{2}+3\sqrt{5})}{50-45}$ $=\displaystyle \frac{25(5\sqrt{2}+3\sqrt{5})}{5}$ $= 5(5\sqrt{2}+3\sqrt{5})$ $= 25\sqrt{2}+15\sqrt{5}$

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