## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$\displaystyle \{\frac{1}{4},9\}$
Substitute $u=\sqrt{y}, u^{2}=y,$ $2u^{2}-7u+3=0$ ... ... factoring, we seek factors of $ac=6$ whose sum is $b=-7$ ... we find $-6$ and $-1$ $2u^{2}-6u-u+3=0$ $2u(u-3)-(u-3)=0$ $(u-3)(2u-1)=0$ $u=3$ or $u=1/2$ ... bring back $y$ $\sqrt{y}=3 \qquad$or$\qquad \displaystyle \sqrt{y}=\frac{1}{2}$ $y=3^{2} \qquad$or$\qquad y=(\displaystyle \frac{1}{2})^{2}$ $y=9 \qquad$or$\qquad y=\displaystyle \frac{1}{4}$ Solution set = $\displaystyle \{\frac{1}{4},9\}$