## College Algebra (10th Edition)

The solution set is{$\frac{3-\sqrt 17}{4} ,\frac{3+\sqrt 17}{4}$}.
$2x^2-3x=1$ $x^2-\frac{3}{2}x=\frac{1}{2}$ Divide each term by 2. $x^2-\frac{3}{2}x+\frac{9}{16}=\frac{1}{2}+\frac{9}{16}$. Add $(\frac{3}{2}\cdot\frac{1}{2})^2=\frac{9}{16}$ to both sides to complete the square. $(x-\frac{3}{4})^2=\frac{17}{16}$ $x-\frac{3}{4}=\pm\sqrt \frac{17}{16}$ Use the square root property and solve. $x=\frac{3}{4}\pm\sqrt \frac{17}{16}$ $x=\frac{3}{4}\pm\frac{\sqrt 17}{4}$ $x=\frac{3}{4}+\frac{\sqrt 17}{4}$, $x=\frac{3}{4}-\frac{\sqrt 17}{4}$ $x=\frac{3+\sqrt 17}{4}$, $x=\frac{3-\sqrt 17}{4}$, The solution set is {$\frac{3-\sqrt 17}{4} ,\frac{3+\sqrt 17}{4}$}.