## College Algebra (11th Edition)

$x=9$
$\bf{\text{Solution Outline:}}$ To solve the given radical equation, $\sqrt{4x}-x+3=0 ,$ use the properties of equality to isolate the radical and then square both sides. 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. $\bf{\text{Solution Details:}}$ Using the properties of equality to isolate the radical, the equation above is equivalent to \begin{array}{l}\require{cancel} \sqrt{4x}-x+3=0 \\\\ \sqrt{4x}=x-3 .\end{array} 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} 4x=(x-3)^2 \\\\ 4x=[(x)^2-2(x)(3)+(3)^2] \\\\ 4x=x^2-6x+9 \\\\ 0=x^2-6x+9-4x \\\\ 0=x^2-10x+9 \end{array} In the trinomial expression above, the value of $ac$ is $1(9)=9$ and the value of $b$ is $-10 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ -1,-9 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} x^2-x-9x+9=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)-(9x-9)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x(x-1)-9(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-9)=0 .\end{array} Equating each factor to zero (Zero Product Property), then \begin{array}{l}\require{cancel} x-1=0 \\\\\text{OR}\\\\ x-9=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x-1=0 \\\\ x=1 \\\\\text{OR}\\\\ x-9=0 \\\\ x=9 .\end{array} Upon checking, only $x=9$ satisfies the original equation.