# How do you solve  1/18x^2 -1/9x-1=0 using the quadratic formula?

Feb 12, 2017

$x = 1 \pm \sqrt{19}$

#### Explanation:

Quadratic formula gives the solution of equation $a {x}^{2} + b x + c = 0$ as $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

In $\frac{1}{18} {x}^{2} - \frac{1}{9} x - 1 = 0$, we have $a = \frac{1}{18}$, $b = - \frac{1}{9}$ and $c = - 1$.

Hence, $x = \frac{- \left(- \frac{1}{9}\right) \pm \sqrt{{\left(- \frac{1}{9}\right)}^{2} - 4 \left(\frac{1}{18}\right) \left(- 1\right)}}{2 \times \frac{1}{18}}$

= $\frac{\frac{1}{9} \pm \sqrt{\frac{1}{81} + \frac{2}{9}}}{\frac{1}{9}}$

= $\frac{\frac{1}{9} \pm \sqrt{\frac{19}{81}}}{\frac{1}{9}}$

= $\left(1 \pm \sqrt{19}\right)$

Alternatively, we could have simplified $\frac{1}{18} {x}^{2} - \frac{1}{9} x - 1 = 0$ by multiplying each term by $18$ to get

${x}^{2} - 2 x - 18 = 0$ and then solution would have been

$x = \frac{- \left(- 2\right) \pm \sqrt{{\left(- 2\right)}^{2} - 4 \times 1 \times \left(- 18\right)}}{2}$

= $\frac{2 \pm \sqrt{4 + 72}}{2} = 1 \pm \sqrt{19}$