## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 7 - Section 7.5 - Multiplying and Dividing Radical Expressions - 7.5 Exercises - Page 475: 25

#### Answer

$26-2\sqrt{105}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given radical expression, $(\sqrt{21}-\sqrt{5})^2 ,$ use the special product on squaring binomials and the properties of radicals. Then combine like terms. $\bf{\text{Solution Details:}}$ 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{21})^2-2(\sqrt{21})(\sqrt{5})+(\sqrt{5})^2 \\\\= 21-2(\sqrt{21})(\sqrt{5})+5 .\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} 21-2\sqrt{21(5)}+5 \\\\= 21-2\sqrt{105}+5 .\end{array} By combining like terms, the expression above is equivalent to \begin{array}{l}\require{cancel} (21+5)-2\sqrt{105} \\\\= 26-2\sqrt{105} .\end{array}

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