# How do you solve the equation x^2-2/3x-26/9=0 by completing the square?

Jan 8, 2017

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

#### Explanation:

Given:

${x}^{2} - \frac{2}{3} x - \frac{26}{9} = 0$

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

Use this with $a = \left(x - \frac{1}{3}\right)$ and $b = \sqrt{3}$ as follows:

$0 = {x}^{2} - \frac{2}{3} x - \frac{26}{9}$

$\textcolor{w h i t e}{0} = {x}^{2} - \frac{2}{3} x + \frac{1}{9} - 3$

$\textcolor{w h i t e}{0} = {\left(x - \frac{1}{3}\right)}^{2} - {\left(\sqrt{3}\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(x - \frac{1}{3}\right) - \sqrt{3}\right) \left(\left(x - \frac{1}{3}\right) + \sqrt{3}\right)$

$\textcolor{w h i t e}{0} = \left(x - \frac{1}{3} - \sqrt{3}\right) \left(x - \frac{1}{3} + \sqrt{3}\right)$

Hence:

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