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

$a=2$
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.