## College Algebra (11th Edition)

$x=\left\{ -27,3 \right\}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $(x^2+24x)^{1/4}=3 ,$ 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} x^2+24x=81 \\\\ x^2+24x-81=0 .\end{array} Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $1(-81)=-81$ and the value of $b$ is $24 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ -3,27 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} x^2-3x+27x-81=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^2-3x)+(27x-81)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x(x-3)+27(x-3)=0 .\end{array} Factoring the $GCF= (x-3)$ of the entire expression above results to \begin{array}{l}\require{cancel} (x-3)(x+27)=0 .\end{array} Equating each factor to zero (Zero Product Property), then \begin{array}{l}\require{cancel} x-3=0 \\\\\text{OR}\\\\ x+27=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x-3=0 \\\\ x=3 \\\\\text{OR}\\\\ x+27=0 \\\\ x=-27 .\end{array} Upon checking, $x=\left\{ -27,3 \right\}$ satisfy the original equation.