How do you complete the square for x^2 -3x?

May 24, 2015

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

To see this, let's multiply out...

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

$= x \left(x - \frac{3}{2}\right) - \frac{3}{2} \left(x - \frac{3}{2}\right)$

$= {x}^{2} - \frac{3}{2} x - \frac{3}{2} x + \frac{9}{4} = {x}^{2} - 3 x + \frac{9}{4}$

Subtract $\frac{9}{4}$ from both sides to get ${x}^{2} - 3 x = {\left(x - \frac{3}{2}\right)}^{2} - \frac{9}{4}$

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

In your case we have $a = 1$, $b = - 3$ and $c = 0$.