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: 50

Answer

$\dfrac{3\sqrt{42}-4\sqrt{15}}{23}$

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}}{3\sqrt{7}+2\sqrt{10}} ,$ is \begin{array}{l}\require{cancel} \dfrac{\sqrt{6}}{3\sqrt{7}+2\sqrt{10}}\cdot\dfrac{3\sqrt{7}-2\sqrt{10}}{3\sqrt{7}-2\sqrt{10}} \\\\= \dfrac{3\sqrt{6(7)}-2\sqrt{6(10)}}{(3\sqrt{7})^2-(2\sqrt{10})^2} \\\\= \dfrac{3\sqrt{42}-2\sqrt{60}}{9(7)-4(10)} \\\\= \dfrac{3\sqrt{42}-2\sqrt{4\cdot15}}{63-40} \\\\= \dfrac{3\sqrt{42}-2\sqrt{(2)^2\cdot15}}{23} \\\\= \dfrac{3\sqrt{42}-2(2)\sqrt{15}}{23} \\\\= \dfrac{3\sqrt{42}-4\sqrt{15}}{23} .\end{array}
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