I've got this inequality:


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:


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.

  • 2
    $\begingroup$ 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

$\dfrac{1}{x} - \dfrac{x}{2x-1} \geq 1 \iff \dfrac{1}{x} - \dfrac{x}{2x-1} - 1 \geq 0 \iff \dfrac{2x-1 - x^2 - x(2x - 1)}{x(2x - 1)} \geq 0 \iff \dfrac{-3x^2 + 3x - 1}{x(2x - 1)} \geq 0$.

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: $0 < x < \dfrac{1}{2}$

  • $\begingroup$ 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

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 $x \in (0,1/2)$.


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


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.

  • $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.