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: 83

Answer

$\dfrac{12-3\sqrt{5}}{11}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To rationalize the given radical expression, $ \dfrac{3}{4+\sqrt{5}} ,$ multiply the numerator and the denominator by the conjugate of the denominator. Then use special products to multiply 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{3}{4+\sqrt{5}}\cdot\dfrac{4-\sqrt{5}}{4-\sqrt{5}} \\\\= \dfrac{3(4-\sqrt{5})}{(4+\sqrt{5})(4-\sqrt{5})} .\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{3(4-\sqrt{5})}{(4)^2-(\sqrt{5})^2} \\\\= \dfrac{3(4-\sqrt{5})}{16-5} \\\\= \dfrac{3(4-\sqrt{5})}{11} .\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{3(4)-3(\sqrt{5})}{11} \\\\= \dfrac{12-3\sqrt{5}}{11} .\end{array}
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