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

Add $\frac{9}{16}$ to both sides of the equation to complete the square. The solutions are $-\frac{3+\sqrt 41}{4}$ and $-\frac{3-\sqrt 41}{4}$, and the solution set is {$-\frac{3+\sqrt 41}{4},-\frac{3-\sqrt 41}{4}$}
$2x^2 + 3x - 4 = 0$ Divide the whole equation by $2$. Thus, the equation becomes: $x^2 + \frac{3}{2}x - 2 = 0$ Add $2$ from both sides to isolate the binomial $x^2 + \frac{3}{2}x$. $x^2 + \frac{3}{2}x - 2 +2 = 0+2$ $x^2 + \frac{3}{2}x = 2$ The coefficient of the x-term is $\frac{3}{2}$. Half of $\frac{3}{2}$ is $\frac{3}{4}$, and $(\frac{3}{4})^2 = \frac{9}{16}$. Thus, add $\frac{9}{16}$ to both sides of the equation to complete the square. $x^2 + \frac{3}{2}x + \frac{9}{16}= 2+\frac{9}{16}$ $x^2 + \frac{3}{2}x + \frac{9}{16}= \frac{41}{16}$ $(x+\frac{3}{4})^2 = \frac{41}{16}$ $x+\frac{3}{4} = \sqrt \frac{41}{16}$ or $x+\frac{3}{4} = -\sqrt \frac{41}{16}$ $x = - \frac{3}{4} +\sqrt \frac{41}{16}$ or $x = - \frac{3}{4} -\sqrt \frac{41}{16}$ Rationalize the denominator: $x = - \frac{3}{4} +\frac{\sqrt 41}{4}$ or $x = - \frac{3}{4} -\frac{\sqrt 41}{4}$ $x = -\frac{3+\sqrt 41}{4}$ or $x = -\frac{3-\sqrt 41}{4}$ The solutions are $-\frac{3+\sqrt 41}{4}$ and $-\frac{3-\sqrt 41}{4}$, and the solution set is {$-\frac{3+\sqrt 41}{4},-\frac{3-\sqrt 41}{4}$}.