Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 11 - Section 11.3 - Geometric Sequences and Series - Exercise Set - Page 858: 123


The answer is $=-3(\sqrt 3+\sqrt 5)$.

Work Step by Step

The given expression is $\frac{6}{\sqrt 3-\sqrt 5}$ Divide and multiply by $\sqrt 3+\sqrt 5$ to rationalize the denominator. $=\frac{6}{\sqrt 3-\sqrt 5}\cdot \frac{\sqrt 3+\sqrt 5}{\sqrt 3+\sqrt 5}$ $=\frac{6(\sqrt 3+\sqrt 5)}{3-5}$ $=\frac{6(\sqrt 3+\sqrt 5)}{-2}$ $=-3(\sqrt 3+\sqrt 5)$.
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