Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 9 - Roots and Radicals - Chapter 9 Review Problem Set - Page 430: 48



Work Step by Step

Multiplying by the conjugate of the denominator and using $(a+b)(a-b)=a^2-b^2$, the simplifed form of given expression, $ \dfrac{\sqrt{6}}{\sqrt{3}-\sqrt{2}} ,$ is \begin{array}{l}\require{cancel} \dfrac{\sqrt{6}}{\sqrt{3}-\sqrt{2}}\cdot\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}} \\\\= \dfrac{\sqrt{6(3)}+\sqrt{6(2)}}{(\sqrt{3})^2-(\sqrt{2})^2} \\\\= \dfrac{\sqrt{18}+\sqrt{12}}{3-2} \\\\= \dfrac{\sqrt{9\cdot2}+\sqrt{4\cdot3}}{3-2} \\\\= \dfrac{\sqrt{(3)^2\cdot2}+\sqrt{(2)^2\cdot3}}{1} \\\\= 3\sqrt{2}+2\sqrt{3} .\end{array}
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