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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!

4 Answers 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 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 this simplifies to $2x^3+8x^2-44x+48>0$, which is true for all positive $x$, so the desired inequality holds for $0. Thus it holds in the region $0, but not outside.

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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, 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, 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.

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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.

Thus for $x>2$, the only solutions to the original inequality are $2.

For $0, we obviously have a solution to the original inequality (since both terms will be positive for $0, and for $x=2$, we have the inequality {2\over\sqrt{14}}>0).

So the solution set is $(0,4)$.

  • 1
    If it weren't obvious that the inequality held for 0, 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...2012-02-23