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: 89


$\displaystyle \frac{3\sqrt{6}+4}{2}$

Work Step by Step

We lose the square roots in the denominator by applying the difference of squares formula: $(a\sqrt{x}+b\sqrt{y})(a\sqrt{x}-b\sqrt{y})=(a\sqrt{x})^{2}-(b\sqrt{y})^{2}=a^{2}x-b^{2}y$ $\displaystyle \frac{5\sqrt{3}-3\sqrt{2}}{3\sqrt{2}-2\sqrt{3}}\color{red}{ \cdot\frac{3\sqrt{2}+2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}} }\qquad$ (rationalize) $=\displaystyle \frac{(5\sqrt{3}-3\sqrt{2})(3\sqrt{2}+2\sqrt{3})}{(3\sqrt{2})^{2}-(2\sqrt{3})^{2}}$ ... use FOIL for the numerator $=\displaystyle \frac{15\sqrt{6}+10\cdot 3-9\cdot 2-6\sqrt{6}}{3^{2}\cdot 2-2^{2}\cdot 3}$ $=\displaystyle \frac{9\sqrt{6}+30-18}{18-12}$ $=\displaystyle \frac{9\sqrt{6}+12}{6}$ $=\displaystyle \frac{3(3\sqrt{6}+4)}{6}$ $=\displaystyle \frac{3\sqrt{6}+4}{2}$
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