Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 9 - Roots and Radicals - Chapters 1-9 Cumulative Review Problem Set - Page 432: 48


$\frac{-6\sqrt 5-2\sqrt 6}{13}$

Work Step by Step

To rationalize the denominator, we multiply both the numerator and the denominator of the expression by $3\sqrt 5+\sqrt 6$ and then simplify: $\frac{-6}{3\sqrt 5-\sqrt 6}\times\frac{3\sqrt 5+\sqrt 6}{3\sqrt 5+\sqrt 6}$ =$\frac{-6(3\sqrt 5+\sqrt 6)}{(3\sqrt 5-\sqrt 6)(3\sqrt 5+\sqrt 6)}$ =$\frac{-18\sqrt 5-6\sqrt 6}{(3\sqrt 5)^{2}-(\sqrt 6)^{2}}$ =$\frac{-18\sqrt 5-6\sqrt 6}{9(5)-6}$ =$\frac{-18\sqrt 5-6\sqrt 6}{45-6}$ =$\frac{3(-6\sqrt 5-2\sqrt 6)}{39}$ =$\frac{-6\sqrt 5-2\sqrt 6}{13}$
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