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: 86


$\displaystyle \frac{8+\sqrt{55}}{3}$

Work Step by Step

We lose the square roots in the denominator by applying the difference of squares formula: $(a\sqrt{x}+b\sqrt{y})(a\sqrt{x}-b\sqrt{y})=(a\sqrt{x})^{2}-(b\sqrt{y})^{2}=a^{2}x-b^{2}y$ $\displaystyle \frac{\sqrt{11}-\sqrt{5}}{\sqrt{11}+\sqrt{5}}\color{red}{ \cdot\frac{\sqrt{11}-\sqrt{5}}{\sqrt{11}-\sqrt{5}} }\qquad$ (rationalize) $=\displaystyle \frac{(\sqrt{11}-\sqrt{5})^{2}}{(\sqrt{11})^{2}-(\sqrt{5})^{2}}$ ... the numerator is a square of a sum, $(A+B)^{2}=A^{2}+2AB+B^{2}$ $=\displaystyle \frac{(\sqrt{11})^{2}+2\sqrt{11}\cdot\sqrt{5}+(\sqrt{5})^{2}}{11-5}$ $=\displaystyle \frac{11+2\sqrt{55}+5}{6}$ $=\displaystyle \frac{16+2\sqrt{55}}{6}$ $=\displaystyle \frac{2(8+\sqrt{55})}{6}$ $=\displaystyle \frac{8+\sqrt{55}}{3}$
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