# How do you solve n^2-4/3n-14/9=0 by completing the square?

Jun 8, 2017

$n = \frac{2}{3} - \sqrt{2}$ or $n = \frac{2}{3} - \sqrt{2}$

#### Explanation:

${n}^{2} - \frac{4}{3} n - \frac{14}{9} = 0$

i.e. ${n}^{2} - 2 \times \frac{2}{3} \times n - \frac{14}{9} = 0$

Now, as ${\left(a - b\right)}^{2} = {a}^{2} - 2 \times a \times b + {b}^{2}$, comparing it with ${n}^{2} - 2 \times \frac{2}{3} \times n$, we find adding ${\left(\frac{2}{3}\right)}^{2} = \frac{4}{9}$, will complete the square and we have

${n}^{2} - 2 \times \frac{2}{3} \times n + {\left(\frac{2}{3}\right)}^{2} - \frac{4}{9} - \frac{14}{9} = 0$

or ${\left(n - \frac{2}{3}\right)}^{2} - \frac{18}{9} = 0$

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

or ${\left(n - \frac{2}{3}\right)}^{2} - {\left(\sqrt{2}\right)}^{2} = 0$ and using ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$

it becomes $\left(n - \frac{2}{3} + \sqrt{2}\right) \left(n - \frac{2}{3} - \sqrt{2}\right) = 0$

Hence either $n - \frac{2}{3} + \sqrt{2} = 0$ i.e. $n = \frac{2}{3} - \sqrt{2}$

or $n - \frac{2}{3} - \sqrt{2} = 0$ i.e. $n = \frac{2}{3} + \sqrt{2}$