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: 80

Answer

$\dfrac{5}{7+3\sqrt{6}}$

Work Step by Step

Multiplying by the conjugate of the numerator, then the rationalized-numerator form of the given expression, $ \dfrac{\sqrt{8}-\sqrt{3}}{\sqrt{2}+\sqrt{3}} ,$ is \begin{array}{l}\require{cancel} \dfrac{\sqrt{8}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}\cdot\dfrac{\sqrt{8}+\sqrt{3}}{\sqrt{8}+\sqrt{3}} \\\\= \dfrac{(\sqrt{8})^2-(\sqrt{3})^2}{\sqrt{2}(\sqrt{8})+\sqrt{2}(\sqrt{3})+\sqrt{3}(\sqrt{8})+\sqrt{3}(\sqrt{3})} \\\\= \dfrac{8-3}{\sqrt{2(8)}+\sqrt{2(3)}+\sqrt{3(8)}+\sqrt{3(3)}} \\\\= \dfrac{5}{\sqrt{16}+\sqrt{6}+\sqrt{24}+\sqrt{9}} \\\\= \dfrac{5}{\sqrt{(4)^2}+\sqrt{6}+\sqrt{4\cdot6}+\sqrt{(3)^2}} \\\\= \dfrac{5}{\sqrt{(4)^2}+\sqrt{6}+\sqrt{(2)^2\cdot6}+\sqrt{(3)^2}} \\\\= \dfrac{5}{4+\sqrt{6}+2\sqrt{6}+3} \\\\= \dfrac{5}{7+3\sqrt{6}} .\end{array}
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