## College Algebra (11th Edition)

$x=-1$
$\bf{\text{Solution Outline:}}$ To solve the given radical equation, $\sqrt{3x+7}=3x+5 ,$ square both sides and then use the properties of equality to express the resulting equation in the form $ax^2+bx+c=0.$ Next is to factor the quadratic equation and solve for the value/s of the variable using the Zero Product Property. Finally, it is a must to do checking of the solution. $\bf{\text{Solution Details:}}$ Squaring both sides and then using the special product on squaring binomials which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 3x+7=(3x+5)^2 \\\\ 3x+7=[(3x)^2+2(3x)(5)+(5)^2] \\\\ 3x+7=9x^2+30x+25 \\\\ -9x^2+(3x-30x)+(7-25)=0 \\\\ -9x^2-27x-18=0 \\\\ \dfrac{-9x^2-27x-18}{-9}=\dfrac{0}{-9} \\\\ x^2+3x+2=0 .\end{array} In the trinomial expression above, the value of $ac$ is $1(2)=2$ and the value of $b$ is $3 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ 1,2 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} x^2+x+2x+2=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)+(2x+2)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x(x+1)+2(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+2)=0 .\end{array} Equating each factor to zero (Zero Product Property), then \begin{array}{l}\require{cancel} x+1=0 \\\\\text{OR}\\\\ x+2=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x+1=0 \\\\ x=-1 \\\\\text{OR}\\\\ x+2=0 \\\\ x=-2 .\end{array} Upon checking, only $x=-1$ satisfies the original equation.