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: 67

Answer

$\dfrac{a+7\sqrt{a}+10}{a-25}$

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{\sqrt{a}+2}{\sqrt{a}-5} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{\sqrt{a}+2}{\sqrt{a}-5}\cdot\dfrac{\sqrt{a}+5}{\sqrt{a}+5} \\\\= \dfrac{\sqrt{a}(\sqrt{a})+\sqrt{a}(5)+2(\sqrt{a})+2(5)}{(\sqrt{a})^2-(5)^2} \\\\= \dfrac{a+5\sqrt{a}+2\sqrt{a}+10}{a-25} \\\\= \dfrac{a+7\sqrt{a}+10}{a-25} .\end{array}
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