# How do you solve using the completing the square method 3x^2-4x-2=0?

Aug 6, 2016

#### Answer:

$x = \frac{- 2 - \sqrt{10}}{3}$ or $x = \frac{- 2 + \sqrt{10}}{3}$

#### Explanation:

Let us divide each term of $3 {x}^{2} - 4 x - 2 = 0$ by $3$, we get

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

Now recalling the identity ${\left(x - a\right)}^{2} = {x}^{2} - 2 a x + {a}^{2}$ and comparing it with ${x}^{2} - \frac{4}{3} x$, we need to add and subtract ${\left(\frac{4}{3 \times 2}\right)}^{2}$ to complete square. Hence ${x}^{2} - \frac{4}{3} x - \frac{2}{3} = 0$ is

$\Leftrightarrow {x}^{2} - 2 \times \frac{4}{6} x + {\left(\frac{4}{6}\right)}^{2} - {\left(\frac{4}{6}\right)}^{2} - \frac{2}{3} = 0$ or

${\left(x + \frac{4}{6}\right)}^{2} - {\left(\frac{2}{3}\right)}^{2} - \frac{2}{3} = 0$ or

${\left(x + \frac{2}{3}\right)}^{2} - \frac{4}{9} - \frac{6}{9} = 0$ or

${\left(x + \frac{2}{3}\right)}^{2} - \frac{10}{9} = 0$ or

${\left(x + \frac{2}{3}\right)}^{2} - {\left(\frac{\sqrt{10}}{3}\right)}^{2} = 0$ or

$\left(x + \frac{2}{3} + \frac{\sqrt{10}}{3}\right) \left(x + \frac{2}{3} - \frac{\sqrt{10}}{3}\right) = 0$

Hence, $x + \frac{2}{3} + \frac{\sqrt{10}}{3} = 0$ or $x + \frac{2}{3} - \frac{\sqrt{10}}{3} = 0$

i.e. $x = \frac{- 2 - \sqrt{10}}{3}$ or $x = \frac{- 2 + \sqrt{10}}{3}$