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 - Review Exercises: Chapter 10: 46

Answer

$a=2$

Work Step by Step

Replacing $x$ with $a$ in $f(x)=\sqrt{x+2}+x$ results to \begin{array}{l}\require{cancel} f(x)=\sqrt{x+2}+x \\\\ f(a)=\sqrt{a+2}+a .\end{array} Substituting $f(a)$ with $4$ and then isolating the radical expression result to \begin{array}{l}\require{cancel} f(a)=\sqrt{a+2}+a \\\\ 4=\sqrt{a+2}+a \\\\ 4-a=\sqrt{a+2} .\end{array} Squaring both sides, the equation above is equivalent to \begin{array}{l}\require{cancel} 4-a=\sqrt{a+2} \\\\ (4-a)^2=\left( \sqrt{a+2} \right)^2 \\\\ (4)^2-2(4)(a)+(a)^2=a+2 \\\\ 16-8a+a^2=a+2 \\\\ a^2-8a-a+16-2=0 \\\\ a^2-9+14=0 .\end{array} Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to\begin{array}{l}\require{cancel} a^2-9+14=0 \\\\ (a-7)(a-2)=0 .\end{array} Equating each factor to zero and then isolating the variable, the solutions are $ a=\{2,7 \}$ If $a=7,$ then \begin{array}{l}\require{cancel} 4=\sqrt{a+2}+a \\\\ 4=\sqrt{7+2}+7 \\\\ 4=\sqrt{9}+7 \\\\ 4=3+7 \\\\ 4=10 \text{ (FALSE)} .\end{array} Hence, only $ a=2 $ satisfies the original equation.
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