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

The solutions are $\frac{7}{2} ± \sqrt\frac{37}{4}$, and the solution set is {${\frac{7}{2}+\sqrt\frac{37}{4}, \frac{7}{2}-\sqrt\frac{37}{4}}$} or {$\frac{7}{2} ± \sqrt\frac{37}{4}$}.
$x^2 - 7x + 3 = 0$ Subtract $3$ from both sides to isolate the binomial $x^2 - 7x$. $x^2 - 7x + 3 -3= 0-3$ $x^2 - 7x = -3$ The coefficient of the x-term is $-7$. Half of $-7$ is $-\frac{7}{2}$, and $(-\frac{7}{2})^2 = \frac{49}{4}$. Thus, add $\frac{49}{4}$ to both sides of the equation to complete the square. $x^2 - 7x + \frac{49}{4}= -3 +\frac{49}{4}$ $x^2 - 7x + \frac{49}{4} = \frac{37}{4}$ $(x-\frac{7}{2})^2 = \frac{37}{4}$ $x -\frac{7}{2} =\sqrt\frac{37}{4}$ or $x -\frac{7}{2} =-\sqrt\frac{37}{4}$ $x = \frac{7}{2}+\sqrt\frac{37}{4}$ or $x = \frac{7}{2}-\sqrt\frac{37}{4}$ The solutions are $\frac{7}{2} ± \sqrt\frac{37}{4}$, and the solution set is {${\frac{7}{2}+\sqrt\frac{37}{4}, \frac{7}{2}-\sqrt\frac{37}{4}}$} or {$\frac{7}{2} ± \sqrt\frac{37}{4}$}.