# How do you solve using completing the square method (-2/3)x^2 + (-4/3)x + 1 = 0?

Jun 6, 2017

$x = - 1 + \sqrt{\frac{5}{2}}$ or $- 1 - \sqrt{\frac{5}{2}}$

#### Explanation:

We have $\left(- \frac{2}{3}\right) {x}^{2} + \left(- \frac{4}{3}\right) x + 1 = 0$

As $\left(- \frac{2}{3}\right) {x}^{2}$ is not a complete square, let us multiply each term by $- \frac{3}{2}$, so that we get ${x}^{2}$, which is a complete square. Then our equation becomes

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

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

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

or ${\left(x + 1\right)}^{2} - \frac{5}{2} = 0$

or ${\left(x + 1\right)}^{2} - {\left(\sqrt{\frac{5}{2}}\right)}^{2} = 0$

i.e. $\left(x + 1 - \sqrt{\frac{5}{2}}\right) \left(x + 1 + \sqrt{\frac{5}{2}}\right) = 0$

Hence, $x = - 1 + \sqrt{\frac{5}{2}}$ or $- 1 - \sqrt{\frac{5}{2}}$