## Intermediate Algebra (12th Edition)

$m=\left\{ 6-\sqrt{3},6+\sqrt{3} \right\}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $(m-6)^2=27 ,$ take the square root of both sides (Square Root Property) and simplify the radical. Then use the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Taking the square root of both sides (Square Root Property), the equation above is equivalent to \begin{array}{l}\require{cancel} m-6=\pm\sqrt{27} .\end{array} Simplifying the radical and using the properties of equality to isolate the variable, the equation above is equivalent to \begin{array}{l}\require{cancel} m-6=\pm\sqrt{9\cdot3} \\\\ m-6=\pm\sqrt{(3)^2\cdot3} \\\\ m-6=\pm3\sqrt{3} \\\\ m=6\pm3\sqrt{3} .\end{array} The solutions are \begin{array}{l}\require{cancel} m=6-\sqrt{3} \\\\\text{OR}\\\\ m=6+\sqrt{3} .\end{array} Hence, $m=\left\{ 6-\sqrt{3},6+\sqrt{3} \right\} .$