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!
 A: 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 $x=4$.) 
A: 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 $0<x<4$.
A: First write the inequality as
$$
{x\sqrt x-(x-2)\sqrt{x+12}\over \sqrt x\sqrt{x+12} }>0
$$
This holds if and only if $x>0$ and $x\sqrt x-(x-2)\sqrt{x+12}>0$.
We have
$$\eqalign{
&x\sqrt x-(x-2)\sqrt{x+12}>0\cr
\iff&x\sqrt x>(x-2)\sqrt{x+12}\cr 

}
$$
For $x>2$, squaring both sides of the above gives 
$$x^3>(x-2)^2 (x+12)$$
or
$$
0>4(2x^2-11x+12)
$$
or
$$
0>(2x-3)(x-4).
$$
The solution to the above is $3/2<x<4$. 
Thus for $x>2$, the only solutions to the original inequality are $2<x<4$.
For $0<x\le2$, we obviously have a solution to the original inequality (since both terms will be positive for $0<x<2$, and for $x=2$, we have the inequality ${2\over\sqrt{14}}>0$).
So the solution set is $(0,4)$.
A: We can combine the two fractions on the left-hand side to get
$$\frac{x\sqrt x-(x-2)\sqrt{x+12}}{\sqrt{x+12}\sqrt{x}}>0$$
and since we need for the denominator to be defined and not $0$, this gives us $x>0$. We can now multiply both sides by $\sqrt{x+12}\sqrt{x}$ to get 
$$x\sqrt x-(x-2)\sqrt{x+12}>0$$
which we rewrite as 
$x\sqrt x>(x-2)\sqrt{x+12}$ and square both sides giving us
$$x^3>\pm(x-2)^2(x+12)=\pm(x^3+8x^2-44x+48)$$
where the $\pm$ is determined by the sign of $x-2$, since if $x\geq 2$ then squaring does not change the sign of either side while if $0<x<2$ then it does.
When $x\geq 2$ this simplifies to $0>8x^2-44x+24$. Applying the quadratic equation, we get that $0=8x^2-44x+48$ at $x=\frac{3}{2},4$ and we can see that between these two values the inequality holds, so we know that our desired inequality holds for $x\geq 2$. When $0<x<2$ this simplifies to $2x^3+8x^2-44x+48>0$, which is true for all positive $x$, so the desired inequality holds for $0<x<2$. Thus it holds in the region $0<x<4$, but not outside.
