# How do I complete the square?

Oct 23, 2015

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

The secret is that $\frac{b}{2 a}$ bit

#### Explanation:

Suppose you are given a quadratic equation to solve:

$2 {x}^{2} - 3 x - 2 = 0$

..which is in the form..

$a {x}^{2} + b x + c = 0$ with $a = 2$, $b = - 3$ and $c = - 2$

$\frac{b}{2 a} = - \frac{3}{4}$

So we find:

$2 {\left(x - \frac{3}{4}\right)}^{2} = 2 \left({x}^{2} - \left(2 \cdot x \cdot \frac{3}{4}\right) + {\left(\frac{3}{4}\right)}^{2}\right)$

$= 2 \left({x}^{2} - \frac{3 x}{2} + \frac{9}{16}\right)$

$= 2 {x}^{2} - 3 x + \frac{9}{8}$

So:

$2 {\left(x - \frac{3}{4}\right)}^{2} - \frac{25}{8} = 2 {\left(x - \frac{3}{4}\right)}^{2} - \frac{9}{8} - 2$

$= 2 {x}^{2} - 3 x + \frac{9}{8} - \frac{9}{8} - 2$

$= 2 {x}^{2} - 3 x - 2$

So:

$2 {x}^{2} - 3 x - 2 = 0$

turns into:

$2 {\left(x - \frac{3}{4}\right)}^{2} - \frac{25}{8} = 0$

Hence:

${\left(x - \frac{3}{4}\right)}^{2} = \frac{25}{16}$

So:

$x - \frac{3}{4} = \pm \sqrt{\frac{25}{16}} = \pm \frac{5}{4}$

and

$x = \frac{3}{4} \pm \frac{5}{4}$