# How do you solve the quadratic equation by completing the square: x^2 – 8x + 12 = 0?

Jul 13, 2015

Work out which factors of 12 add together to give 8. You can then factorise the equation. 12 is either $3 \cdot 4$ or $6 \cdot 2$ - $3 + 4 = 7$ so that won't work, but $6 + 2 = 8$ so that does work.

#### Explanation:

$\left(x - 6\right) \left(x - 2\right) = 0$

Jul 13, 2015

$x = 6$ or $x = 2$
$\textcolor{w h i t e}{\text{XXXX}}$(solved by completion of squares method)

#### Explanation:

Given x^2–8x+12=0

$\textcolor{w h i t e}{\text{XXXX}}$Move the constant to the right side
${x}^{2} - 8 x = - 12$
$\textcolor{w h i t e}{\text{XXXX}}$If ${x}^{2}$ and $- 8 x$ are the first two terms of a squared binomial:
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$${\left(x + a\right)}^{2} = {x}^{2} + 2 a x + {a}^{2}$
$\textcolor{w h i t e}{\text{XXXX}}$then the third term needs to be $\left({a}^{2}\right) = {\left(\frac{8}{2}\right)}^{2}$
${x}^{2} - 8 x + {\left(\frac{8}{2}\right)}^{2} = - 12 + {\left(\frac{8}{2}\right)}^{2}$

${x}^{2} - 8 x + {4}^{2} = - 12 + 16$

$\textcolor{w h i t e}{\text{XXXX}}$rewrite the left side as a squared binomial
$\textcolor{w h i t e}{\text{XXXX}}$ and simplify the right side.
${\left(x - 4\right)}^{2} = 4$

$\textcolor{w h i t e}{\text{XXXX}}$Take the square root of both sides
$x - 4 = \pm 2$

$\textcolor{w h i t e}{\text{XXXX}}$Add 4 to both sides
$x = 6$
or
$x = 2$