Question 890100
{{{8x-12 > 8/x}}}


Do NOT multiply each side by x.  


{{{18x-12-8/x>0}}}
{{{9x-6-4/x>0}}}
Bring the first two terms to higher-terms according to denominator x.
{{{(9x^2-6x-4)/x>0}}}


There, you want to use the NUMERATOR to see how it is related to zero; the denominator will not really be relevant, except that the solution for x must EXCLUDE x=0.


The important critical points are at x being the zeros of the numerator.


{{{x=(6+- sqrt(36+4*9*4))/(2*9)}}}
{{{x=(6+- sqrt(36+36*4))/(2*9)}}}
{{{x=(6+- 6*sqrt(5))/(6*3)}}}
{{{x=1/3+- sqrt(5)/3}}}


The inequality will be satisfied for {{{highlight(x<1/3-(1/3)sqrt(5))}}} or for {{{highlight(x>1/3+(1/3)sqrt(5))}}}----
The possibility of x=0 will no longer be necessary to consider since it is not within the domain for this inequality.