# How do you solve by completing the square: x^2 - 8x - 1 = 0?

Mar 30, 2015

If ${x}^{2} - 8 x$ are the first 2 terms of a squared expression of the form ${\left(x + a\right)}^{2}$
then the third term should be
${\left(\frac{- 8}{2}\right)}^{2} = 16$

If we are going to replace
${x}^{2} - 8 x - 1$
with
${x}^{2} - 8 x + 16$
we will need to add $+ 17$ to both sides of the equation

${x}^{2} - 8 x + 16 = 17$

re-write the left-side as a square
${\left(x - 4\right)}^{2} = 17$

then take the square root of both sides
$x - 4 = \pm \sqrt{17}$

$\rightarrow x = 4 + \sqrt{17}$ or $x = 4 - \sqrt{17}$