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

Question

11383 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-05-30T11:50:59

In general the square ${(x + a)}^{2} = x^{2} + 2 a x + a^{2}$ $(x+a)^2 = x^2+2ax+a^2$

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

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

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

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

$x - 4 = \pm \sqrt{3}$ $x-4 = +-sqrt(3)$

$x = 4 \pm \sqrt{3}$ $x = 4+-sqrt(3)$

How do you complete the square to solve x2−8x+13=0x^2 - 8x + 13 = 0?