# How do you solve the quadratic equation by completing the square: x^2 -12x-2=0?

Jul 13, 2015

$x = 6 + \sqrt{38}$ or $x = 6 - \sqrt{38}$
$\textcolor{w h i t e}{\text{XXXX}}$(using the completing the squares method)

#### Explanation:

Given ${x}^{2} - 12 x - 2 = 0$

Completing the square is simpler if we move the constant to the right side:
$\textcolor{w h i t e}{\text{XXXX}}$${x}^{2} - 12 x = 2$

If ${x}^{2}$ and $- 12 x$ are the first two terms of a squared binomial expansion:
$\textcolor{w h i t e}{\text{XXXX}}$${\left(x - a\right)}^{2} = \left({x}^{2} - 2 a x + {a}^{2}\right)$
then
$\textcolor{w h i t e}{\text{XXXX}}$$a = 6$
and
we need to add ${a}^{2} = 36$ to both sides to complete the square

$\textcolor{w h i t e}{\text{XXXX}}$${x}^{2} - 12 x + 36 = 2 + 36$

Rewriting as a squared binomial (and simplifying the right side)
$\textcolor{w h i t e}{\text{XXXX}}$${\left(x - 6\right)}^{2} = 38$

Taking the square root of both sides
$\textcolor{w h i t e}{\text{XXXX}}$$x - 6 = \pm \sqrt{38}$

Adding $6$ to both sides:
$\textcolor{w h i t e}{\text{XXXX}}$$x = 6 \pm \sqrt{38}$