# How do you complete the square for : x^2 + 6x?

May 16, 2015

${\left(x + 3\right)}^{2} = {x}^{2} + 6 x + 9$

So ${x}^{2} + 6 x = {\left(x + 3\right)}^{2} - 9$

If you were actually trying to solve ${x}^{2} + 6 x = 0$, then completing the square would be a round about way of doing it, but would look something like this:

$0 = {x}^{2} + 6 x = {x}^{2} + 6 x + 9 - 9 = {\left(x + 3\right)}^{2} - 9$

Adding 9 to both sides you get ${\left(x + 3\right)}^{2} = 9$

So $x + 3 = \pm \sqrt{9} = \pm 3$

Then subtracting 3 from both sides, you get

$x = - 3 \pm 3$

In other words $x = 0$ or $x = - 6$.