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I've got this inequality:

$$\dfrac{1}{x}-\dfrac{x}{2x-1}\geq1$$

And the solutions are supposed to be $(0,1/2)$. But why? Whenever I make my inequality table, it ends up telling me the solution is $(-∞,0) \cup (1/2,+∞)$

Can someone help me understand what is going on? I really need to know this to study for a test.

EDIT: Alright, to make it clearer this is what I've got after simplifying everything:

$$\dfrac{\left(-3x^2+3x-1\right)}{x\left(2x-1\right)}>=0$$

After that I just put all the values on a table, in which I look for the numbers that make 2x-1 and x equal 0. The polynomial in the numerator can never be zero since its discriminant is less than 0. That's basically where I'm stuck at. I would post the table, but I've no clue where I could create one, I'm doing this on paper.

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    $\begingroup$ Detail out why you think the solution is what you say it is. Try out a value in each interval, say $x=-1, \frac14, 1$. $\endgroup$ – Macavity May 16 '14 at 3:47
  • $\begingroup$ So after you get that $x = 0$ and $x=\frac12$ are the only points where the LHS can change signs, you should find what the signs are in each of the intervals $(-\infty, 0), (0, \frac12), (\frac12, \infty)$. Once you find the right interval(s) where the inequality holds, you need to just check the end points to verify if they should be included / excluded. $\endgroup$ – Macavity May 16 '14 at 4:02
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$\dfrac{1}{x} - \dfrac{x}{2x-1} \geq 1 \iff \dfrac{1}{x} - \dfrac{x}{2x-1} - 1 \geq 0 \iff \dfrac{2x-1 - x^2 - x(2x - 1)}{x(2x - 1)} \geq 0 \iff \dfrac{-3x^2 + 3x - 1}{x(2x - 1)} \geq 0$.

Observe that: $-3x^2 + 3x - 1 < 0$, $\forall x$ because $\triangle = 3^2 - 4(-3)(-1) = 9 - 12 = -3 < 0$, and $-3 < 0$. Thus the inequality is equivalent to:

$x(2x - 1) < 0$, and this gives: $0 < x < \dfrac{1}{2}$

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  • $\begingroup$ I think I understand now, that polynomial is always less than zero, because of its negative discriminant, nonetheless, does that mean that whenever I've got a polynomial with a negative discriminant, then I should change the sign or something? $\endgroup$ – Argus May 16 '14 at 4:00
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The inequality is equivalent to $$ \begin{array}{rclr} \frac{2x-1 - x^2}{x(2x-1)} -1 &\geq& 0 & \iff \\ \frac{2x-1 - x^2-2x^2 + x}{x(2x-1)} &\geq& 0 & \iff \\ \frac{-3x^2 + 3x -1 }{x(2x-1)} &\geq& 0. & \ \\ \end{array} $$ The numerator is a negative term (the polynomial opens downwards and has no real roots) thus the quotient is non-negative precisely when $x(2x-1) < 0$, i.e. when $x \in (0,1/2)$.

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Combine the fractions on the LHS of the inequality by finding the common denominator and note that the numerator is then the square of $x-1$:

$$\frac{1}{x}-\frac{x}{2x-1}=\frac{2x-1-x^2}{x(2x-1)}=-\frac{(x-1)^2}{x(2x-1)}=\frac{(x-1)^2}{x(1-2x)}.$$

Assuming $x\neq 1$, we divide both sides by $(x-1)^2$:

$$1\leq \frac{(x-1)^2}{x(1-2x)} \implies \frac{1}{(x-1)^2}\leq \frac{1}{x(1-2x)}$$

This tells us that $1/x$ and $1/(1-2x)$ must either be both positive or both negative, and hence that $x$ and $1-2x$ must either be both positive or negative as well. All that's left now is some very simple case work.

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  • $\begingroup$ @Macavity I fail to see how. $\frac{1}{x}-\frac{x}{2x-1}=\frac{1\cdot(2x-1)}{x(2x-1)}-\frac{x^2}{x(2x-1)}$ $\endgroup$ – David H May 16 '14 at 4:04

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