# 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..

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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 $(0,4)$.

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If it weren't obvious that the inequality held for $0<x\le2$, then you should follow the method used in Andre Nicolas' answer. Come to think of it, as I used sort of ad-hoc method, you should just use Andre's method... – David Mitra Feb 23 '12 at 23:56

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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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 $0<x<4$.

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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.

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