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

Jul 11, 2017

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

#### Explanation:

$\text{require the coefficient of " x^2" term to be unity}$

$\Rightarrow 3 \left({x}^{2} - \frac{2}{3} x - 4\right) = 0$

$\text{to complete the square}$

add (1/2"coefficient of the x-term")^2" to " x^2-2/3x

$3 \left({x}^{2} - \frac{2}{3} x \textcolor{red}{+ \frac{1}{9}} \textcolor{m a \ge n t a}{- \frac{1}{9}} - 4\right) = 0$

$\text{we must also subtract } \frac{1}{9}$

$\Rightarrow 3 \left({\left(x - \frac{1}{3}\right)}^{2} - \frac{37}{9}\right) = 0$

$\text{distributing gives}$

$3 {\left(x - \frac{1}{3}\right)}^{2} - \frac{37}{3} = 0$

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

$\textcolor{b l u e}{\text{take the square root of both sides}}$

$\Rightarrow x - \frac{1}{3} = \pm \sqrt{\frac{37}{9}} \leftarrow \text{ note plus or minus}$

$\text{add "1/3" to both sides}$

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