Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 7 - Section 7.5 - Multiplying with More Than One Term and Rationalizing Denominators - Exercise Set - Page 550: 101

Answer

$\displaystyle \frac{1}{\sqrt{x+5}+\sqrt{x}}$

Work Step by Step

We lose the square roots in the numerator by applying the difference of squares formula: $(a\sqrt{x}+b\sqrt{y})(a\sqrt{x}-b\sqrt{y})=(a\sqrt{x})^{2}-(b\sqrt{y})^{2}$ $=a^{2}x-b^{2}y$ $\displaystyle \frac{\sqrt{x+5}-\sqrt{x}}{5}\color{red}{ \cdot\frac{\sqrt{x+5}+\sqrt{x}}{\sqrt{x+5}+\sqrt{x}} }\qquad$ (rationalize) $=\displaystyle \frac{(\sqrt{x+5})^{2}-(\sqrt{x})^{2}}{5(\sqrt{x+5}+\sqrt{x})}$ $=\displaystyle \frac{x+5-x}{5(\sqrt{x+5}+\sqrt{x})}$ $=\displaystyle \frac{5}{5(\sqrt{x+5}+\sqrt{x})}$ $=\displaystyle \frac{1}{\sqrt{x+5}+\sqrt{x}}$
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