# How do you complete the square to solve x^2 - 8x + 13 = 0?

May 30, 2015

In general the square ${\left(x + a\right)}^{2} = {x}^{2} + 2 a x + {a}^{2}$

If $\left(- 8 x\right)$ is the second term of a square of this form then
$a = - 4$ and ${a}^{2} = 16$

${x}^{2} - 8 x + 13 = 0$

${\left(x - 4\right)}^{2} - 16 + 13 = 0$

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

$x - 4 = \pm \sqrt{3}$

$x = 4 \pm \sqrt{3}$