# How do you solve \frac { 1} { x ^ { 2} } - \frac { 1} { x } = 6?

$x = - \frac{1}{2} , \frac{1}{3}$

#### Explanation:

We need the fractions to have a common denominator. We can get there through creative use of the number 1:

$\frac{1}{x} ^ 2 - \frac{1}{x} = 6$

$\frac{1}{x} ^ 2 \left(1\right) - \frac{1}{x} \left(1\right) = 6 \left(1\right)$

$\frac{1}{x} ^ 2 \left(\frac{1}{1}\right) - \frac{1}{x} \left(\frac{x}{x}\right) = 6 \left({x}^{2} / {x}^{2}\right)$

$\frac{1}{x} ^ 2 - \frac{x}{x} ^ 2 = \frac{6 {x}^{2}}{x} ^ 2$

And now we can multiply through by ${x}^{2}$ which will eliminate the denominators, giving:

$1 - x = 6 {x}^{2}$

$6 {x}^{2} + x - 1 = 0$

I'll use the quadratic formula to solve.

$x = \frac{- b \setminus \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

and we have $a = 6 , b = 1 , c = - 1$

$x = \frac{- 1 \setminus \pm \sqrt{{1}^{2} - 4 \left(6\right) \left(- 1\right)}}{2 \left(6\right)}$

$x = \frac{- 1 \setminus \pm \sqrt{25}}{12} = \frac{- 1 \pm 5}{12}$

$\therefore x = \frac{- 1 + 5}{12} = \frac{4}{12} = \frac{1}{3}$
$\therefore x = \frac{- 1 - 5}{12} = - \frac{6}{12} = - \frac{1}{2}$

Here's the graph:

graph{6x^2+x-1[-1,1,-2,2]}