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: 36

Answer

$\dfrac{\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}}=3-2\sqrt{2}+2\sqrt{3}-\sqrt{6}$

Work Step by Step

$\dfrac{\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}}$ Multiply the numerator and the denominator of this expression by the conjugate of the denominator and simplify if possible: $\dfrac{\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}}=\dfrac{\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}}\cdot\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}=\dfrac{(\sqrt{3}+\sqrt{4})(\sqrt{2}-\sqrt{3})}{(\sqrt{2})^{2}-(\sqrt{3})^{2}}=...$ $...=\dfrac{\sqrt{6}-\sqrt{3^{2}}+\sqrt{8}-\sqrt{12}}{2-3}=\dfrac{\sqrt{6}-3+2\sqrt{2}-2\sqrt{3}}{-1}=...$ $...=3-2\sqrt{2}+2\sqrt{3}-\sqrt{6}$
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