# How do you solve by completing the square x^2- 4x-11=0?

Apr 1, 2015

$x = 2 - \sqrt{15} , 2 + \sqrt{15}$

We should reform this equation by a full square. So we have:
${x}^{2} - 4 x + c = 0$

I don't want to go deeper in this, lets conclude this part with $c = 4$

So:

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

Now, lets put ${\left(x - 2\right)}^{2}$ in the given equation.

$G i v e n :$ 1) ${x}^{2} - 4 x - 11 = 0$
${\left(x - 2\right)}^{2} + n = {x}^{2} - 4 x - 11$
$n = - 15$

So:

${\left(x - 2\right)}^{2} - 15 = {x}^{2} - 4 x - 11$
${\left(x - 2\right)}^{2} - 15 = 0$
$\sqrt{{\left(x - 2\right)}^{2}} = 15$
$\left\mid x - 2 \right\mid = \sqrt{15}$
$x - 2 = - \sqrt{15} , \sqrt{15}$
$x = 2 - \sqrt{15} , 2 + \sqrt{15}$