## Intermediate Algebra (12th Edition)

$x=0$
$\bf{\text{Solution Outline:}}$ To solve the given radical equation, $\sqrt{9-x}=x+3 ,$ square both sides of the equation. Use special products and concepts of solving quadratic equations. Finally, do checking of the solution with the original equation. $\bf{\text{Solution Details:}}$ Squaring both sides of the equation results to \begin{array}{l}\require{cancel} \left( \sqrt{9-x} \right)^2=\left( x+3 \right)^2 \\\\ 9-x=\left( x+3 \right)^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 expression above is equivalent to \begin{array}{l}\require{cancel} 9-x=(x)^2+2(x)(3)+(3)^2 \\\\ 9-x=x^2+6x+9 .\end{array} In the form $ax^2+bx+c=0,$ the equation above is equivalent to \begin{array}{l}\require{cancel} -x^2+(-x-6x)+(9-9)=0 \\\\ -x^2-7x=0 \\\\ -1(-x^2-7x)=-1(0) \\\\ x^2+7x=0 .\end{array} Using factoring by $GCF,$ the equation above is equivalent to \begin{array}{l}\require{cancel} x(x+7)=0 .\end{array} Equating each factor to zero (Zero Product Property), the solutions are \begin{array}{l}\require{cancel} x=0 \\\\\text{OR}\\\\ x+7=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x=0 \\\\\text{OR}\\\\ x+7=0 \\\\ x=-7 .\end{array} Upon checking, only $x=0$ satisfies the original equation.