# How do you solve 3x^2 – 24x + 24 = 0 using completing the square?

Jul 3, 2015

Logical first step: divide everything by $3$

#### Explanation:

$\to {x}^{2} - 8 x + 8 = 0$

Completing the square would mean: taking half of the $x$-coefficient and squaring that:
${x}^{2} - 8 x + {\left(- 4\right)}^{2} = {x}^{2} - 8 x + 16 = {\left(x - 4\right)}^{2}$

But we still have to balance the $16$ with the $8$ we had:
$\to {x}^{2} - 8 x + 16 - 8 = 0$
$\to {\left(x - 4\right)}^{2} - 8 = 0 \to {\left(x - 4\right)}^{2} = 8$

So $x - 4 = \sqrt{8} = 2 \sqrt{2} \to x = 4 + 2 \sqrt{2}$
Or $x - 4 = - \sqrt{8} = - 2 \sqrt{2} \to x = 4 - 2 \sqrt{2}$

Often written as ${x}_{1 , 2} = 4 \pm 2 \sqrt{2}$