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: 47


$4\sqrt 3-4\sqrt 2$

Work Step by Step

To rationalize the denominator, we multiply both the numerator and the denominator of the expression by $\sqrt 3-\sqrt 2$ and then simplify: $\frac{4}{\sqrt 3+\sqrt 2}\times\frac{\sqrt 3-\sqrt 2}{\sqrt 3-\sqrt 2}$ =$\frac{4(\sqrt 3-\sqrt 2)}{(\sqrt 3+\sqrt 2)(\sqrt 3-\sqrt 2)}$ =$\frac{4\sqrt 3-4\sqrt 2}{(\sqrt 3)^{2}-(\sqrt 2)^{2}}$ =$\frac{4\sqrt 3-4\sqrt 2}{3-2}$ =$\frac{4\sqrt 3-4\sqrt 2}{1}$ =$4\sqrt 3-4\sqrt 2$
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