Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Section 7.5 - Rationalizing Denominators and Numerators of Radical Expressions - Exercise Set - Page 445: 45

Answer

$-5+2\sqrt{6}$

Work Step by Step

Rationalizing the denominator of $ \dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}} $ results to \begin{array}{l} \dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}} \cdot \dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}} \\\\= \dfrac{(\sqrt{2}-\sqrt{3})^2}{(\sqrt{2})^2-(\sqrt{3})^2} \\\\= \dfrac{(\sqrt{2})^2+2(\sqrt{2})(-\sqrt{3})+(-\sqrt{3})^2}{2-3} \\\\= \dfrac{2-2\sqrt{6}+3}{-1} \\\\= -2+2\sqrt{6}-3 \\\\= -5+2\sqrt{6} .\end{array}
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