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: 85


$ 4+\sqrt{15}$

Work Step by Step

We lose the square roots in the denominator by applying the difference of squares formula: $(a\sqrt{x}+b\sqrt{y})(a\sqrt{x}-b\sqrt{y})=(a\sqrt{x})^{2}-(b\sqrt{y})^{2}=a^{2}x-b^{2}y$ $\displaystyle \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}color{red}{ \cdot\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}} }\qquad$ (rationalize) $=\displaystyle \frac{(\sqrt{5}+\sqrt{3})^{2}}{(\sqrt{5})^{2}-(\sqrt{3})^{2}}$ ... the numerator is a square of a sum, $(A+B)^{2}=A^{2}+2AB+B^{2}$ $=\displaystyle \frac{(\sqrt{5})^{2}+2\sqrt{5}\cdot\sqrt{3}+(\sqrt{3})^{2}}{5-3}$ $=\displaystyle \frac{5+2\sqrt{15}+3}{2}$ $=\displaystyle \frac{8+2\sqrt{15}}{2}$ $=\displaystyle \frac{2(4+\sqrt{15})}{2}$ $=4+\sqrt{15}$
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