<s>[INST] 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!
 [/INST] Note that our expression is only defined when $x>0$. Bring to a common denominator $\sqrt{x+12}\sqrt{x}$. We get
$$\frac{x\sqrt{x}-(x-2)\sqrt{x+12}}{\sqrt{x+12}\sqrt{x}}.$$
The bottom is safely positive, so we want to find out where 
$$x\sqrt{x}-(x-2)\sqrt{x+12}>0.$$
This expression can only change sign when we travel across points where the expression is $0$. So we solve
$$x\sqrt{x}-(x-2)\sqrt{x+12}=0.$$
To find out where this could happen, we bring the negative stuff to the other side, and then square both sides. We are looking at
$$x^3-(x-2)^2(x+12)=0.$$
Expand. The $x^3$ terms  cancel, and we  get a quadratic. Solve. 
The solutions should turn out to be $x=3/2$ and $x=4$. This divides the region we are interested in into parts $(0,3/2)$, $(3/2,4)$ and $(4,\infty)$. We also need to worry a tiny bit about $3/2$ and  $4$.
Now look at either our original function, or $g(x)=x\sqrt{x}-(x-2)\sqrt{x+12}$. Evaluate it at convenient "test points" in our intervals.
For example, to deal with $(0,3/2)$, we can use the test point $x=1$. It is easy to see that $g(1)$ is positive. Now look at a convenient test point in $(3/2,4)$, like $x=2$. Clearly, $g(2)$ is positive. Finally, deal with $(4,\infty)$. We may need a calculator. Let $x=9$. We find that $g(9)$ is negative. So there is a change of sign only at $x=4$. For $0<x<4$, our expression is $>0$.  For $x\ge 4$, our expression is $\le 0$. (It is exactly $0$ at $x=4$.) 
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