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: 87

Answer

$\dfrac{6\sqrt{5}-4\sqrt{3}}{33} $

Work Step by Step

$\bf{\text{Solution Outline:}}$ To rationalize the given radical expression, $ \dfrac{2}{3\sqrt{5}+2\sqrt{3}} ,$ 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{2}{3\sqrt{5}+2\sqrt{3}} \cdot\dfrac{3\sqrt{5}-2\sqrt{3}}{3\sqrt{5}-2\sqrt{3}} \\\\= \dfrac{2(3\sqrt{5}-2\sqrt{3})}{(3\sqrt{5}+2\sqrt{3})(3\sqrt{5}-2\sqrt{3})} .\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{2(3\sqrt{5}-2\sqrt{3})}{(3\sqrt{5})^2-(2\sqrt{3})^2} \\\\= \dfrac{2(3\sqrt{5}-2\sqrt{3})}{9(5)-4(3)} \\\\= \dfrac{2(3\sqrt{5}-2\sqrt{3})}{45-12} \\\\= \dfrac{2(3\sqrt{5}-2\sqrt{3})}{33} .\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{2(3\sqrt{5})+2(-2\sqrt{3})}{33} \\\\= \dfrac{6\sqrt{5}-4\sqrt{3}}{33} .\end{array}
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