## College Algebra (11th Edition)

$x=\left\{ -2,-1 \right\}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $\sqrt{x+2}-x=2 ,$ isolate the radical expression. Then square both sides and express in the form $ax^2+bx+c=0.$ Use concepts of solving quadratic equations to find the values of $x$. Finally, do checking of the solutions with the original equation. $\bf{\text{Solution Details:}}$ Using the properties of equality, the given equation is equivalent to \begin{array}{l}\require{cancel} \sqrt{x+2}=x+2 .\end{array} Squaring both sides of the given equation results to \begin{array}{l}\require{cancel} x+2=(x+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 equation above is equivalent to \begin{array}{l}\require{cancel} x+2=(x)^2+2(x)(2)+(2)^2 \\\\ x+2=x^2+4x+4 .\end{array} In the form $ax^2+bx+c=0,$ the equation above is equivalent to \begin{array}{l}\require{cancel} -x^2+(x-4x)+(2-4)=0 \\\\ -x^2-3x-2=0 \\\\ -1(-x^2-3x-2)=-1(0) \\\\ x^2+3x+2=0 .\end{array} Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the factored form of the equation above is \begin{array}{l}\require{cancel} (x+2)(x+1)=0 .\end{array} Equating each factor to zero (Zero Product Property) results to \begin{array}{l}\require{cancel} x+2=0 \\\\\text{OR}\\\\ x+1=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x+2=0 \\\\ x=-2 \\\\\text{OR}\\\\ x+1=0 \\\\ x=-1 .\end{array} Upon checking, $x=\left\{ -2,-1 \right\}$ satisfy the original equation.