## College Algebra (11th Edition)

$x=3$
$\bf{\text{Solution Outline:}}$ To solve the given radical equation, $x-\sqrt{2x+3}=0 ,$ use the properties of equality to isolate the radical. Then square both sides and use concepts of quadratic equations to solve for the variable. Finally, it is a must to do checking of the solution. $\bf{\text{Solution Details:}}$ Using properties of equality to isolate the radical, the equation above is equivalent to \begin{array}{l}\require{cancel} x=\sqrt{2x+3} .\end{array} Squaring both sides and using properties of equality to express the result in $ax^2+bx+c=0$ form result to \begin{array}{l}\require{cancel} x^2=2x+3 \\\\ x^2-2x-3=0 .\end{array} In the trinomial expression above, the value of $ac$ is $1(-3)=-3$ and the value of $b$ is $-2 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ 1,-3 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} x^2+x-3x-3=0 .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (x^2+x)-(3x+3)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x(x+1)-3(x+1)=0 .\end{array} Factoring the $GCF= (x+1)$ of the entire expression above results to \begin{array}{l}\require{cancel} (x+1)(x-3)=0 .\end{array} Equating each factor to zero (Zero Product Property), then \begin{array}{l}\require{cancel} x+1=0 \\\\\text{OR}\\\\ x-3=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x+1=0 \\\\ x=-1 \\\\\text{OR}\\\\ x-3=0 \\\\ x=3 .\end{array} Upon checking, only $x=3$ satisfies the original equation.