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 551: 102

Answer

$ \displaystyle \frac{1}{\sqrt{x+7}+\sqrt{x}}$

Work Step by Step

We lose the square roots in the numerator by applying the difference of squares formula: $(a\sqrt{x}+b\sqrt{y})(a\sqrt{x}-b\sqrt{y})=(a\sqrt{x})^{2}-(b\sqrt{y})^{2}$ $=a^{2}x-b^{2}y$ $\displaystyle \frac{\sqrt{x+7}-\sqrt{x}}{7}\color{red}{ \cdot\frac{\sqrt{x+7}+\sqrt{x}}{\sqrt{x+7}+\sqrt{x}} }\qquad$ (rationalize) $=\displaystyle \frac{(\sqrt{x+7})^{2}-(\sqrt{x})^{2}}{7(\sqrt{x+7}+\sqrt{x})}$ $=\displaystyle \frac{x+7-x}{7(\sqrt{x+7}+\sqrt{x})}$ $=\displaystyle \frac{7}{7(\sqrt{x+7}+\sqrt{x})}$ $=\displaystyle \frac{1}{\sqrt{x+7}+\sqrt{x}}$
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