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

Dec 2, 2016

${x}^{2} - \frac{2}{3} x - \frac{26}{9} = \left(x - \frac{1}{3} + \sqrt{3}\right) \left(x - \frac{1}{3} - \sqrt{3}\right)$

#### Explanation:

This is a polynomial and not an equation and hence cannot be solved for $x$. However, we can factorize it using completing the square, as follows.

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

= $\underline{{x}^{2} - 2 \times \left(\frac{1}{3}\right) \times x + {\left(\frac{1}{3}\right)}^{2}} - {\left(\frac{1}{3}\right)}^{2} - \frac{26}{9}$

= ${\left(x - \frac{1}{3}\right)}^{2} - \frac{1}{9} - \frac{26}{9}$

= ${\left(x - \frac{1}{3}\right)}^{2} - \frac{27}{9}$

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

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

and as ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$ hence above is equal to

= $\left(x - \frac{1}{3} + \sqrt{3}\right) \left(x - \frac{1}{3} - \sqrt{3}\right)$