Answer
$\{-3,3\}$
Work Step by Step
Apply rule if $|u|=0$ then $u=0$ to obtain:
$x^2-9 =0$
$x^2-3^2 =0$
Use special rule $a^2-b^2=(a+b)(a-b)$.
$(x+3)(x-3) =0$
Use Zero-Product Property which states that if $ab=0$, then $a=0$ or $b=0$ or both are $0$ to obtain:
$x+3=0\quad$ or $\quad x-3 =0$
Solve each equation for $x$.
$x=-3\quad $ or $\quad x=3$
Hence, the solution set of the given equation is $\{-3,3\}$.
