Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 7 - Section 7.5 - Multiplying and Dividing Radical Expressions - 7.5 Exercises: 97

Answer

$\dfrac{10k-5\sqrt{kq}}{4k-q}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To rationalize the given radical expression, $ \dfrac{5\sqrt{k}}{2\sqrt{k}+\sqrt{q}} ,$ multiply the numerator and the denominator by the conjugate of the denominator. Then use special products to simplify the result. $\bf{\text{Solution Details:}}$ Multiplying the numerator and the denominator of the given expression by the conjugate of the denominator, the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{5\sqrt{k}}{2\sqrt{k}+\sqrt{q}}\cdot\dfrac{2\sqrt{k}-\sqrt{q}}{2\sqrt{k}-\sqrt{q}} \\\\= \dfrac{5\sqrt{k}(2\sqrt{k}-\sqrt{q})}{(2\sqrt{k}+\sqrt{q})(2\sqrt{k}-\sqrt{q})} .\end{array} Using the product of the sum and difference of like terms which is given by $(a+b)(a-b)=a^2-b^2,$ the expression above is equivalent \begin{array}{l}\require{cancel} \dfrac{5\sqrt{k}(2\sqrt{k}-\sqrt{q})}{(2\sqrt{k})^2-(\sqrt{q})^2} \\\\= \dfrac{5\sqrt{k}(2\sqrt{k}-\sqrt{q})}{4k-q} .\end{array} Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{5(2)(\sqrt{k})^2-5(\sqrt{k})(\sqrt{q})}{4k-q} \\\\= \dfrac{10k-5(\sqrt{k})(\sqrt{q})}{4k-q} .\end{array} Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{10k-5\sqrt{k(q)}}{4k-q} \\\\= \dfrac{10k-5\sqrt{kq}}{4k-q} .\end{array}
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