Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 10 - Exponents and Radicals - Study Summary - Practice Exercises: 12

Answer

$\dfrac{45-15\sqrt{5}}{4}$

Work Step by Step

Multiplying by the conjugate of the denominator, then \begin{array}{l}\require{cancel} \dfrac{15}{3+\sqrt{5}} \\\\= \dfrac{15}{3+\sqrt{5}}\cdot\dfrac{3-\sqrt{5}}{3-\sqrt{5}} \\\\= \dfrac{15(3-\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})} .\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{15(3)+15(-\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})} \\\\= \dfrac{45-15\sqrt{5}}{(3+\sqrt{5})(3-\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{45-15\sqrt{5}}{(3+\sqrt{5})(3-\sqrt{5})} \\\\= \dfrac{45-15\sqrt{5}}{(3)^2-(\sqrt{5})^2} \\\\= \dfrac{45-15\sqrt{5}}{9-5} \\\\= \dfrac{45-15\sqrt{5}}{4} .\end{array}
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