College Algebra (10th Edition)

$\left\{-5, 5\right\}$
Write 25 as $5^2$ to obtain: $x^2-5^2=0$ Factor the difference of two squares using the formula $a^2-b^2=(a-b)(a+b)$ to obtain: $(x-5)(x+5)=0$ Use the Zero-Product Property by equating each factor to zero then solving equation to obtain: $\begin{array}{ccc} &x-5=0 &\text{ or } &x+5=0 \\&x=5 &\text{ or } &x=-5 \end{array}$ Thus, the solution set is $\left\{-5, 5\right\}$.