## Algebra and Trigonometry 10th Edition

$x=\pm 9$
$f(x)=x^2-81$ Since we are solving for all the real values of $x$ when $f(x) = 0$, set the function $f(x)=x^2-81$ equal to zero and we get the following equation: $x^2-81=0$ Now add 81 to both sides of the equation so that all of the constants are on the right side of the equation and all of the $x$ terms are on the left side. $x^2-81+81=0+81$ $x^2=81$ Once all of the constants are on the right and all the $x$ terms are on the left, solve for $x$ by taking the square root of both sides. $\sqrt{x^2}=\sqrt{81}$ Note: Remember taking the square root of both sides of the equation results in both a positive and negative answer. This is why the $\pm$ appears in the line below. $x=\pm\sqrt{81}$ $x=\pm 9$