How do you solve by completing the square: x^2 +8x = 7?

Mar 30, 2015

Since ${\left(x + a\right)}^{2} = {x}^{2} + 2 a x + {a}^{2}$

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

Add $16$ to each side of the equation, giving
${x}^{2} + 8 x + 16 = 23$
which is the same as
${\left(x + 4\right)}^{2} = 23$

Taking the square root of both sides
$x + 4 = \pm \sqrt{23}$

$x = - 4 + \sqrt{23}$
or
$x = - 4 - \sqrt{23}$