Elementary Algebra

The solution set is $\left\{-\frac{5}{9}, \frac{5}{9}\right\}$.
Add $-25$ to both sides to find: $81x^2-25=0 \\(9x)^2-5^2=0$ RECALL: A difference of two squares can be factored using the formula: $a^2-b^2=(a-b)(a+b)$ Factor the difference of two squares to find: $(9x-5)(9x+5)=0$ Equate each factor to 0 then solve each equation to find: $9x-5=0 \text{ or } 9x+5=0\\ 9x=5 \text{ or } 9x=-5 \\x=\frac{5}{9} \text{ or } x=-\frac{5}{9}$ The solution set is $\left\{-\frac{5}{9}, \frac{5}{9}\right\}$.