Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 10 - Section 10.5 - Rationalizing Numerators and Denominators of Radical Expressions - Exercise Set: 35

Answer

$\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}=2\sqrt{6}-5$

Work Step by Step

$\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}$ Multiply the numerator and the denominator of this expression by the conjugate of the denominator and simplify if possible: $\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}=\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}\cdot\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}=\dfrac{(\sqrt{2}-\sqrt{3})^{2}}{(\sqrt{2})^{2}-(\sqrt{3})^{2}}=...$ $...=\dfrac{(\sqrt{2})^{2}-2(\sqrt{2})(\sqrt{3})+(\sqrt{3})^{2}}{2-3}=\dfrac{2-2\sqrt{6}+3}{-1}=...$ $...=\dfrac{5-2\sqrt{6}}{-1}=2\sqrt{6}-5$
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