Question 890100
how do I solve this inequality?

8x-12 > 8/x
<pre>
{{{8x - 12 > 8/x}}}........{{{x <> 0}}}
{{{8x^2 - 12x > 8}}} ------ Multiplying both sides be LCD, x
{{{8x^2 - 12x - 8 > 0}}}
{{{4(2x^2 - 3x - 2) > 4(0)}}}
{{{2x^2 - 3x - 2 > 0}}}
{{{(2x + 1)(x - 2) > 0}}} 
2x + 1 = 0          OR          x - 2 = 0 ------- Converting inequality to equality
2x = - 1            OR          x = 0 + 2
{{{x = - 1/2}}}              OR          x = 2
Critical points are {{{- 1/2}}}, {{{0}}}, since {{{x <> 0}}}, and {{{2}}}.
Therefore the four intervals to check are: {{{x < - 1/2}}}, {{{- 1/2 < x < 0}}}, {{{0 < x < 2}}}, and {{{x > 2}}}.
Intervals that satisfy INEQUALITY: {{{highlight_green(highlight_green(- 1/2 < x < 0))}}} and {{{highlight_green(highlight_green(x > 0))}}}