## Calculus with Applications (10th Edition)

$-3$ and $3$
Divide both sides of the equation by 4: $4x^{2}-36=0\qquad.../\div 4$ $x^{2}-9=0$ ... recognize a difference of squares (of x and 3) $A^{2}-B^{2}=(A+B)(A-B)$ $(x+3)(x-3)=0$ By the zero product principle, $x+3=0$ or $x-3=0$ $x=-3$ or $x=3$ The solutions are $-3$ and $3$.