Solving the inequality $\frac{x}{\sqrt{x+12}} - \frac{x-2}{\sqrt{x}} > 0$ I'm having troubles to solve the following inequality..
$$\frac{x}{\sqrt{x+12}} - \frac{x-2}{\sqrt{x}} > 0$$
I know that the result is $x>0$ and $x<4$ but I cannot find a way to the result..
Thanks in advance!

The domain of possible values of $x$ is $(0,+\infty)$. Rewrite this inequality as
$$
\frac{x}{\sqrt{x+12}}>\frac{x-2}{\sqrt{x}}
$$
If $0<x<2$, then LHS is poistive and RHS is negative and the inequality holds for $x\in (0,2)$. If $x\geq2$ both sides are positive, so you can square them and get
$$
\frac{x^2}{x+12}>\frac{(x-2)^2}{x}
$$
After some simplificatoins you will get the following
$$
-\frac{4(2x^2-11x+12)}{x(x+12)}>0\Longleftrightarrow \frac{2x^2-11x+12}{x(x+12)}<0\Longleftrightarrow\frac{(2x-3)(x-4)}{x(x+12)}<0
$$
The solution of the last inequality is $x\in (-12,0)\cup(3/2,4)$. But we are considering case $x\geq 2$, so $x\in[2,4)$.
After union of results of two cases we get $0<x<4$.
