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: 87


$\displaystyle \frac{x-2\sqrt{x}-3}{x-9}$

Work Step by Step

We lose the square roots in the denominator by applying the difference of squares formula: $(a\sqrt{x}+b\sqrt{y})(a\sqrt{x}-b\sqrt{y})=(a\sqrt{x})^{2}-(b\sqrt{y})^{2}=a^{2}x-b^{2}y$ $\displaystyle \frac{\sqrt{x}+1}{\sqrt{x}+3}\color{red}{ \cdot\frac{\sqrt{x}-3}{\sqrt{x}-3} }\qquad$ (rationalize) $=\displaystyle \frac{(\sqrt{x}+1)(\sqrt{x}-3)}{(\sqrt{x})^{2}-(3)^{2}}$ ... use FOIL for the numerator $=\displaystyle \frac{(\sqrt{x})^{2}-3\sqrt{x}+\sqrt{x}-3}{x-9}$ $=\displaystyle \frac{x-2\sqrt{x}-3}{x-9}$
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