## College Algebra (11th Edition)

$x=-1$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $\sqrt{2x+3}=x+2 ,$ square both sides and then 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. $\bf{\text{Solution Details:}}$ Squaring both sides of the given equation results to \begin{array}{l}\require{cancel} 2x+3=(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} 2x+3=(x)^2+2(x)(2)+(2)^2 \\\\ 2x+3=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+(2x-4x)+(3-4)=0 \\\\ -x^2-2x-1=0 \\\\ -1(-x^2-2x-1)=-1(0) \\\\ x^2+2x+1=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+1)(x+1)=0 \\\\ (x+1)^2=0 .\end{array} Taking the square root of both sides results to \begin{array}{l}\require{cancel} x+1=\pm\sqrt{0} \\\\ x+1=0 \\\\ x=-1 .\end{array} Upon checking, $x=-1$ satisfies the original equation.