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

Question

1223 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-03T01:23:40

$x=-1/2,1/3$

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

$1/x^2-1/x=6$

$1/x^2(1)-1/x(1)=6(1)$

$1/x^2(1/1)-1/x(x/x)=6(x^2/x^2)$

$1/x^2-x/x^2=(6x^2)/x^2$

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

$1-x=6x^2$

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

I'll use the quadratic formula to solve.

$x = (-b \pm sqrt(b^2-4ac)) / (2a)$

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

$x = (-1 \pm sqrt(1^2-4(6)(-1))) / (2(6))$

$x = (-1 \pm sqrt25) / 12=(-1pm5)/12$

$:. x=(-1+5)/12=4/12=1/3$
$:. x=(-1-5)/12=-6/12=-1/2$

Here's the graph:

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

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