## Precalculus (6th Edition) Blitzer

Solution set = $\displaystyle \{0,\frac{8}{3}\}$
Use the square root property: If $u^{2}=d$ then $u=\pm\sqrt{d}$ Here, $u=3x-4, \quad d=16,\quad\sqrt{d}=4$ Apply the property $3x-4=\pm 4 \quad$ ... add 4 to both sides $3x=4\pm 4\quad$ ... divide with 3 $x=\displaystyle \dfrac{4\pm 4}{3}\Rightarrow\left\{\begin{array}{l} x=\dfrac{4-4}{3}=0,\\ \\ x=\dfrac{4+4}{3}=\dfrac{8}{3} \end{array}\right.$ Solution set = $\displaystyle \{0,\frac{8}{3}\}$