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

Published by Pearson

Chapter 10 - Exponents and Radicals - Review Exercises: Chapter 10 - Page 693: 41

Answer

$-4\sqrt{10}+4\sqrt{15}$

Work Step by Step

Multiplying by the conjugate of the denominator, the given expression is equivalent to \begin{array}{l}\require{cancel} \dfrac{4\sqrt{5}}{\sqrt{2}+\sqrt{3}} \\\\= \dfrac{4\sqrt{5}}{\sqrt{2}+\sqrt{3}}\cdot\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}} \\\\= \dfrac{4\sqrt{5}(\sqrt{2}-\sqrt{3})}{(\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3})} .\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{4\sqrt{5}(\sqrt{2})+4\sqrt{5}(-\sqrt{3})}{(\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3})} \\\\= \dfrac{4\sqrt{5(2)}-4\sqrt{5(3)}}{(\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3})} \\\\= \dfrac{4\sqrt{10}-4\sqrt{15}}{(\sqrt{2}+\sqrt{3})(\sqrt{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 to \begin{array}{l}\require{cancel} \dfrac{4\sqrt{10}-4\sqrt{15}}{(\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3})} \\\\= \dfrac{4\sqrt{10}-4\sqrt{15}}{(\sqrt{2})^2-(\sqrt{3})^2} \\\\= \dfrac{4\sqrt{10}-4\sqrt{15}}{2-3} \\\\= \dfrac{4\sqrt{10}-4\sqrt{15}}{-1} \\\\= -4\sqrt{10}+4\sqrt{15} .\end{array}

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