# How do you solve 0=3x^2-2x-12 by completing the square?

May 16, 2015

In general $a {x}^{2} + b x + c = a {\left(x + \frac{b}{2 a}\right)}^{2} + \left(c - {b}^{2} / \left(4 a\right)\right)$

For your problem, $a = 3$, $b = - 2$ and $c = - 12$, so we get

$0 = a {x}^{2} + b x + c = 3 {x}^{2} - 2 x - 12$

$= 3 {\left(x + \frac{- 2}{2 \cdot 3}\right)}^{2} + \left(- 12 - {\left(- 2\right)}^{2} / \left(4 \cdot 3\right)\right)$

$= 3 {\left(x - \frac{2}{6}\right)}^{2} - \left(12 + \frac{4}{12}\right)$

$= 3 {\left(x - \frac{1}{3}\right)}^{2} - \left(12 + \frac{1}{3}\right)$

$= 3 {\left(x - \frac{1}{3}\right)}^{2} - \frac{37}{3}$

Adding $\frac{37}{3}$ to both sides we get

$3 {\left(x - \frac{1}{3}\right)}^{2} = \frac{37}{3}$

Dividing both sides by 3 we get

${\left(x - \frac{1}{3}\right)}^{2} = \frac{37}{{3}^{2}}$

Hence

$x - \frac{1}{3} = \pm \sqrt{\frac{37}{3} ^ 2} = \pm \frac{\sqrt{37}}{3}$

Adding 1/3 to both sides, we get

$x = \frac{1}{3} \pm \frac{\sqrt{37}}{3} = \frac{1 \pm \sqrt{37}}{3}$

May 16, 2015

If $3 {x}^{2} - 2 x - 12 = 0$
then
${x}^{2} - \frac{2}{3} x - 4 = 0$

isolate the constant
${x}^{2} - \frac{2}{3} x = 4$

complete the square
${x}^{2} - \frac{2}{3} x + {\left(\frac{1}{3}\right)}^{2} = 4 + {\left(\frac{1}{3}\right)}^{2}$

${\left(x - \frac{1}{3}\right)}^{2} = \frac{37}{9}$

take the square root
$x - \frac{1}{3} = \pm \frac{\sqrt{37}}{3}$

simplify
$x = \frac{1 \pm \sqrt{37}}{3}$