# How do you solve by completing the square for x^2 - 5x = 9?

Jun 14, 2015

${\left(x - \frac{5}{2}\right)}^{2} = {x}^{2} - 5 x + \frac{25}{4} = 9 + \frac{25}{4} = \frac{61}{4}$

Hence: $x = \frac{5}{2} \pm \frac{\sqrt{61}}{2}$

#### Explanation:

${\left(x - \frac{5}{2}\right)}^{2}$

$= {x}^{2} - 5 x + \frac{25}{4}$

$= 9 + \frac{25}{4}$

$= \frac{36}{4} + \frac{25}{4}$

$= \frac{36 + 25}{4}$

$= \frac{61}{4}$

So:

$x - \frac{5}{2} = \pm \sqrt{\frac{61}{4}} = \pm \frac{\sqrt{61}}{\sqrt{4}} = \pm \frac{\sqrt{61}}{2}$

Add $\frac{5}{2}$ to both ends to get:

$x = \frac{5}{2} \pm \frac{\sqrt{61}}{2}$

In general,

$a {x}^{2} + b x + c = a {\left(x + \frac{b}{2 a}\right)}^{2} + \left(c - {b}^{2} / \left(4 a\right)\right)$