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: 90


$\displaystyle \frac{41+5\sqrt{30}}{49}$

Work Step by Step

We lose the square roots in the denominator by applying the difference of squares formula: $(a\sqrt{x}+b\sqrt{y})(a\sqrt{x}-b\sqrt{y})=(a\sqrt{x})^{2}-(b\sqrt{y})^{2}=a^{2}x-b^{2}y$ $\displaystyle \frac{2\sqrt{6}+\sqrt{5}}{3\sqrt{6}-\sqrt{5}}\color{red}{ \cdot\frac{3\sqrt{6}+\sqrt{5}}{3\sqrt{6}+\sqrt{5}} }\qquad$ (rationalize) $=\displaystyle \frac{(2\sqrt{6}+\sqrt{5})(3\sqrt{6}+\sqrt{5})}{(3\sqrt{6})^{2}-(\sqrt{5})^{2}}$ ... use FOIL for the numerator $=\displaystyle \frac{6\cdot 6+2\sqrt{30}+3\sqrt{30}+5}{3^{2}\cdot 6-5}$ $=\displaystyle \frac{36+5\sqrt{30}+5}{54-5}$ $=\displaystyle \frac{41+5\sqrt{30}}{49}$
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