# How do you solve by completing the square for 2x^2 + 2x = 0?

May 16, 2015

It's a rather strange way to solve it as it is easy to see that $x = 0$ and $x = - 1$ are both solutions. However, here's how you would do it...

$0 = 2 {x}^{2} + 2 x = 2 {\left(x + \frac{1}{2}\right)}^{2} - \frac{1}{2}$

Add 1/2 to both sides to get

$2 {\left(x + \frac{1}{2}\right)}^{2} = \frac{1}{2}$

Divide both sides by 2 to get

${\left(x + \frac{1}{2}\right)}^{2} = \frac{1}{4}$

So

$x + \frac{1}{2} = \pm \sqrt{\frac{1}{4}} = \pm \frac{\sqrt{1}}{\sqrt{{2}^{2}}} = \pm \frac{1}{2}$

Subtract 1/2 from both sides to get

$x = - \frac{1}{2} \pm \frac{1}{2} = \frac{- 1 \pm 1}{2}$

May 17, 2015

If $2 {x}^{2} + 2 x = 0$, then ${x}^{2} + x = 0$.

To complete this square, I need to square half of the coefficient of the 'x', which is 1/2.

${\left(\frac{1}{2}\right)}^{2} = \frac{1}{4}$ , so ${x}^{2} + x + \frac{1}{4} = \frac{1}{4}$

${\left(x + \frac{1}{2}\right)}^{2} = \frac{1}{4}$

$| x + \frac{1}{2} | = \sqrt{\frac{1}{4}}$ (square roots can be positive or negative)

$| x + \frac{1}{2} | = \frac{1}{2}$

$x = \left\{- 1 , 0\right\}$