## Intermediate Algebra (12th Edition)

$t=\left\{ \dfrac{1-2\sqrt{6}}{4},\dfrac{1+2\sqrt{6}}{4} \right\}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $(1-4p)^2=24 ,$ 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} 1-4p=\pm\sqrt{24} .\end{array} Simplifying the radical and then using the properties of equality to isolate the variable, the equation above is equivalent to \begin{array}{l}\require{cancel} 1-4p=\pm\sqrt{4\cdot6} \\\\ 1-4p=\pm\sqrt{(2)^2\cdot6} \\\\ 1-4p=\pm2\sqrt{6} \\\\ -4p=-1\pm2\sqrt{6} \\\\ p=\dfrac{-1\pm2\sqrt{6}}{-4} \\\\ p=\dfrac{1\pm2\sqrt{6}}{4} .\end{array} The solutions are \begin{array}{l}\require{cancel} p=\dfrac{1-2\sqrt{6}}{4} \\\\\text{OR}\\\\ p=\dfrac{1+2\sqrt{6}}{4} .\end{array} Hence, $t=\left\{ \dfrac{1-2\sqrt{6}}{4},\dfrac{1+2\sqrt{6}}{4} \right\} .$