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: 42

Answer

$4-2\sqrt{10}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given radical expression, $ [(\sqrt{5}-\sqrt{2})-\sqrt{3}][(\sqrt{5}-\sqrt{2})+\sqrt{3}] ,$ use the special products on multiplying the sum and difference of like terms and squaring binomials. Then, combine like terms. $\bf{\text{Solution Details:}}$ 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} (\sqrt{5}-\sqrt{2})^2-(\sqrt{3})^2 \\\\= (\sqrt{5}-\sqrt{2})^2-3 .\end{array} Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to \begin{array}{l}\require{cancel} [(\sqrt{5})^2-2(\sqrt{5})(\sqrt{2})+(\sqrt{2})^2-3 \\\\= 5-2(\sqrt{5})(\sqrt{2})+2-3 .\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} 5-2\sqrt{5(2)}+2-3 \\\\= 5-2\sqrt{10}+2-3 .\end{array} Combining like terms, the expression above is equivalent to \begin{array}{l}\require{cancel} (5+2-3)-2\sqrt{10} \\\\= 4-2\sqrt{10} .\end{array}
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