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

The solutions are $2b$ and $-b$, and solution set is {$2b, -b$}.
$x^2 - bx = 2b^2$. Solve for $x$. The coefficient of the x-term is $-b$. Half of $-b$ is $-\frac{b}{2}$, and $(-\frac{b}{2})^2$ is $\frac{b^2}{4}$. Add $\frac{b^2}{4}$ to both sides of the equation to complete the square. $x^2 - bx +\frac{b^2}{4}= 2b^2+\frac{b^2}{4}$ $x^2 - bx +\frac{b^2}{4}= \frac{9b^2}{4}$ $(x-\frac{b}{2})^2 = \frac{9b^2}{4}$ $x-\frac{b}{2}= ±\sqrt\frac{9b^2}{4}$ $x = \frac{b}{2}+\frac{3b}{2}$ or $\frac{b}{2}-\frac{3b}{2}$ $x = 2b$ or $x=-b$ The solutions are $2b$ and $-b$, and solution set is {$2b, -b$}.