Solve the following inequality I have the following inequality $\frac {2x^2}{x+2} < x-2$. I tried to solve it the with the following steps.
step 1 $\frac {2x^2}{x+2} < x-2$
step 2 $\frac {2x^2}{x+2} - (x-2) < 0$
step 3 $\frac {2x^2}{x+2} - \frac {(x-2)(x+2)}{1(x+2)} < 0$
step 4 $\frac {2x^2}{x+2} - \frac {x^2-2^2}{x+2} < 0$
step 5 $\frac {2x^2 - x^2 + 4}{x+2} < 0$
step 5 $\frac {x^2 + 4}{x+2} < 0$
step 6 I used character study to get the result x > 2. But this is incorrect.
Where did I go wrong with this?. I feel that I made a mistake somewhere in step 2 but not sure what I did wrong.
Thanks!

$\left(\frac{x^2+4}{ x+2}\right) < 0$ implies $\left(\frac{1}{x+2}\right) < 0$ because $x^2 + 4 >0$ which implies $x+2 < 0$ and then $x < -2$ Please note $\left(\frac{1}{x+2}\right) = 0$ can never occur.
