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: 77


$\displaystyle \frac{13\sqrt{11}+39}{2}$

Work Step by Step

We lose the square roots in the denominator by applying the difference of squares formula: $(a\sqrt{x}+b\sqrt{y})(a\sqrt{x}-b\sqrt{y})=(a\sqrt{x})^{2}-(b\sqrt{y})^{2}=a^{2}x-b^{2}y$ $\displaystyle \frac{13}{\sqrt{11}-3} \displaystyle \color{red}{ \cdot\frac{\sqrt{11}+3}{\sqrt{11}+3} }\qquad$ (rationalize) $ =\displaystyle \frac{13(\sqrt{11}+3)}{(\sqrt{11})^{2}-3^{2}} =\frac{13(\sqrt{11}+31)}{11-9} =\frac{13(\sqrt{11}+3)}{2}$ = $\displaystyle \frac{13\sqrt{11}+39}{2}$
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