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 - Test: Chapter 10: 8

Answer

$f(5+\sqrt{2})=27+10\sqrt{2}$

Work Step by Step

Substituting $x$ with $(5+\sqrt{2})$ in $f(x)=x^2,$ then \begin{array}{l}\require{cancel} f(x)=x^2 \\\\ f(5+\sqrt{2})=(5+\sqrt{2})^2 .\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} f(5+\sqrt{2})=(5)^2+2(5)(\sqrt{2})+(\sqrt{2})^2 \\\\ f(5+\sqrt{2})=25+10\sqrt{2}+2 \\\\ f(5+\sqrt{2})=27+10\sqrt{2} .\end{array}
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