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

Published by Pearson

# Chapter 10 - Exponents and Radicals - Review Exercises: Chapter 10 - Page 693: 39

#### Answer

$f(2-\sqrt{a})=4-4\sqrt{a}+a$

#### Work Step by Step

Substituting $x$ with $(2-\sqrt{a})$ in $f(x)=x^2,$ results to \begin{array}{l}\require{cancel} f(x)=x^2 \\\\ f(2-\sqrt{a})=(2-\sqrt{a})^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(2-\sqrt{a})=(2-\sqrt{a})^2 \\\\ f(2-\sqrt{a})=(2)^2-2(2)(\sqrt{a})+(\sqrt{a})^2 \\\\ f(2-\sqrt{a})=4-4\sqrt{a}+a .\end{array}

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