Elementary Algebra

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

Chapter 9 - Roots and Radicals - 9.4 - Products and Quotients Involving Radicals - Problem Set 9.4: 58

Answer

$15+5\sqrt{7}$

Work Step by Step

Multiplying by the conjugate of the denominator and using $(a+b)(a-b)=a^2-b^2,$ the given expression, $ \dfrac{10}{3-\sqrt{7}} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{10}{3-\sqrt{7}}\cdot\dfrac{3+\sqrt{7}}{3+\sqrt{7}} \\\\= \dfrac{10(3+\sqrt{7})}{(3)^2-(\sqrt{7})^2} \\\\= \dfrac{10(3+\sqrt{7})}{9-7} \\\\= \dfrac{10(3+\sqrt{7})}{2} \\\\= \dfrac{\cancel{2}(5)(3+\sqrt{7})}{\cancel{2}} \\\\= 5(3+\sqrt{7}) \\\\= 15+5\sqrt{7} .\end{array}
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