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: 90

Answer

$\sqrt{15}+\sqrt{10}+3\sqrt{2}+2\sqrt{3}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To rationalize the given radical expression, $ \dfrac{\sqrt{5}+\sqrt{6}}{\sqrt{3}-\sqrt{2}} ,$ 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{\sqrt{5}+\sqrt{6}}{\sqrt{3}-\sqrt{2}} \cdot\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}} \\\\= \dfrac{(\sqrt{5}+\sqrt{6})(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})} .\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{(\sqrt{5}+\sqrt{6})(\sqrt{3}+\sqrt{2})}{(\sqrt{3})^2-(\sqrt{2})^2} \\\\= \dfrac{(\sqrt{5}+\sqrt{6})(\sqrt{3}+\sqrt{2})}{3-2} \\\\= \dfrac{(\sqrt{5}+\sqrt{6})(\sqrt{3}+\sqrt{2})}{1} \\\\= (\sqrt{5}+\sqrt{6})(\sqrt{3}+\sqrt{2}) .\end{array} Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to\begin{array}{l}\require{cancel} \sqrt{5}(\sqrt{3})+\sqrt{5}(\sqrt{2})+\sqrt{6}(\sqrt{3})+\sqrt{6}(\sqrt{2}) .\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} \sqrt{5(3)}+\sqrt{5(2)}+\sqrt{6(3)}+\sqrt{6(2)} \\\\= \sqrt{15}+\sqrt{10}+\sqrt{18}+\sqrt{12} .\end{array} Writing the radicand as an expression that contains a factor that is a perfect power of the index and extracting the root of that factor result to \begin{array}{l}\require{cancel} \sqrt{15}+\sqrt{10}+\sqrt{9\cdot2}+\sqrt{4\cdot3} \\\\= \sqrt{15}+\sqrt{10}+\sqrt{(3)^2\cdot2}+\sqrt{(2)^2\cdot3} \\\\= \sqrt{15}+\sqrt{10}+3\sqrt{2}+2\sqrt{3} .\end{array}
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