## Precalculus (6th Edition)

$\color{blue}{\left\{-27, 3\right\}}$
Raise both sides to the 4th power to obtain: $((x^2+24x)^{1/4})^4=3^4 \\x^2+24x=81 \\x^2+24x-81=0$ Factor the trinomial to obtain: $(x+27)(x-3)=0$ Use the Zero-Factor Property by equating each factor to zero, then solve each equation to obtain: \begin{array}{ccc} &x+27=0 &\text{or} &x-3=0 \\&x=-27 &\text{or} &x=3 \end{array} Upon checking, both proposed solutions satisfy the original equation. Thus, the solution set is $\color{blue}{\left\{-27, 3\right\}}$.