## College Algebra (11th Edition)

$x=\left\{ -\dfrac{64}{3},4 \right\}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $(3x^2+52x)^{1/4}=4 ,$ raise both sides to the exponent equal to $4 .$ Then express the resulting equation in the form $ax^2+bx+c=0,$ and use the concepts of factoring quadratic equations to solve the resulting equation. Finally, do checking if the solution satisfies the original equation. $\bf{\text{Solution Details:}}$ Raising both sides to the fourth power, the given equation becomes \begin{array}{l}\require{cancel} 3x^2+52x=256 \\\\ 3x^2+52x-256=0 .\end{array} Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $3(-256)=-768$ and the value of $b$ is $52 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ -12,64 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 3x^2-12x+64x-256=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} (3x^2-12x)+(64x-256)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 3x(x-4)+64(x-4)=0 .\end{array} Factoring the $GCF= (x-4)$ of the entire expression above results to \begin{array}{l}\require{cancel} (x-4)(3x+64)=0 .\end{array} Equating each factor to zero (Zero Product Property), then \begin{array}{l}\require{cancel} x-4=0 \\\\\text{OR}\\\\ 3x+64=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x-4=0 \\\\ x=4 \\\\\text{OR}\\\\ 3x+64=0 \\\\ 3x=-64 \\\\ x=-\dfrac{64}{3} .\end{array} Upon checking, $x=\left\{ -\dfrac{64}{3},4 \right\}$ satisfy the original equation.