## Intermediate Algebra for College Students (7th Edition)

Add $\frac{25}{16}$ to both sides of the equation to complete the square. The solutions are $\frac{5}{4}±\frac{\sqrt31}{4}i$, and the solution set is {$\frac{5}{4}+\frac{\sqrt31}{4}i, \frac{5}{4}-\frac{\sqrt31}{4}i$}.
$2x^2 - 5x + 7 = 0$ Divide the whole equation by $2$. Thus, the equation becomes: $x^2 - \frac{5}{2}x +\frac{7}{2} = 0$ Subtract $\frac{7}{2}$ from both sides to isolate the binomial $x^2 - \frac{5}{2}x$. $x^2 - \frac{5}{2}x +\frac{7}{2} - \frac{7}{2} = 0 - \frac{7}{2}$ $x^2 - \frac{5}{2}x = - \frac{7}{2}$ The coefficient of the x-term is $- \frac{5}{2}$. Half of $- \frac{5}{2}$ is $- \frac{5}{4}$, and $(- \frac{5}{4})^2 = \frac{25}{16}$. Thus, add $\frac{25}{16}$ to both sides of the equation to complete the square. $x^2 - \frac{5}{2}x +\frac{25}{16}= - \frac{7}{2} +\frac{25}{16}$ $x^2 - \frac{5}{2}x +\frac{25}{16}= - \frac{31}{16}$ $(x-\frac{5}{4})^2 = - \frac{31}{16}$ $x-\frac{5}{4} = \sqrt {- \frac{31}{16}}$ or $x-\frac{5}{4} = -\sqrt {- \frac{31}{16}}$ $x = \frac{5}{4} +\sqrt {- \frac{31}{16}}$ or $x =\frac{5}{4} -\sqrt {- \frac{31}{16}}$ $x = \frac{5}{4}+\sqrt \frac{31}{16}\sqrt {-1}$ or $x = \frac{5}{4} -\sqrt \frac{31}{16}\sqrt {-1}$ $x =\frac{5}{4}+\frac{\sqrt{31}}{4}i$ or $x = \frac{5}{4}-\frac{\sqrt{31}}{4}i$ The solutions are $\frac{5}{4}±\frac{\sqrt{31}}{4}i$, and the solution set is {$\frac{5}{4}+\frac{\sqrt{31}}{4}i, \frac{5}{4}-\frac{\sqrt{31}}{4}i$}.