## Intermediate Algebra (12th Edition)

$x=\left\{ -5-4\sqrt{3},-5+4\sqrt{3} \right\}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $(t+5)^2=48 ,$ 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} t+5=\pm\sqrt{48} .\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} t+5=\pm\sqrt{16\cdot3} \\\\ t+5=\pm\sqrt{(4)^2\cdot3} \\\\ t+5=\pm4\sqrt{3} \\\\ t=-5\pm4\sqrt{3} .\end{array} The solutions are \begin{array}{l}\require{cancel} t=-5-4\sqrt{3} \\\\\text{OR}\\\\ t=-5+4\sqrt{3} .\end{array} Hence, $x=\left\{ -5-4\sqrt{3},-5+4\sqrt{3} \right\} .$