## College Algebra (10th Edition)

The solution set is $\left\{-3, 3\right\}$
Write 9 as $3^2$ to obtain: $x^2-3^2=0$ Factor the difference of two squares using the formula $a^2-b^2=(a-b)(a+b)$ to obtain: $(x-3)(x+3)=0$ Use the Zero-Product Property by equating each factor to zero and then solving the equation to obtain: $\begin{array}{ccc} &x-3=0 &\text{ or } &x+3=0 \\&x=3 &\text{ or } &x=-3 \end{array}$ Thus, the solution set is $\left\{-3, 3\right\}$.