# How do you complete the square to solve -x² +6x +9=0?

May 31, 2015

$- {x}^{2} + 6 x + 9 = 0$
is equivalent to
$\textcolor{w h i t e}{\text{XXXXX}}$${x}^{2} - 6 x - 9 =$

to help keep thing simple, move the constant to the right side as
$\textcolor{w h i t e}{\text{XXXXX}}$${x}^{2} - 6 x = 9$

in the general form of the squared binomial
$\textcolor{w h i t e}{\text{XXXXX}}$
${\left(x + a\right)}^{2} = {x}^{2} + 2 a x + {a}^{2}$
so if ${x}^{2} - 6 x$ are the first two terms of a squared binomial, $a = 3$

and to complete the square, we need to add an extra
$\textcolor{w h i t e}{\text{XXXXX}}$${a}^{2} = {\left(- 3\right)}^{2} = 9$

Therefore we write
$\textcolor{w h i t e}{\text{XXXXX}}$${x}^{2} - 6 x + 9 = 9 + 9$

$\textcolor{w h i t e}{\text{XXXXX}}$${\left(x - 3\right)}^{2} = 18$

Taking the square root of both sides:
$\textcolor{w h i t e}{\text{XXXXX}}$$x - 3 = \pm 3 \sqrt{2}$

and
$\textcolor{w h i t e}{\text{XXXXX}}$$x = 3 + 3 \sqrt{2}$ or $x = 3 - 3 \sqrt{2}$