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

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Answer 1 · 2015-05-30T11:50:59

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

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

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

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

$(x-4)^2 = 3$

$x-4 = +-sqrt(3)$

$x = 4+-sqrt(3)$

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