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

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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]}