Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This worksheet

This question:

$$w^2 - w \leq 0$$

This answer:

$$(-\infty, -1] \cup [0, 1]$$

Isn't this wrong ? At $w = -2$, it becomes: $(-2)^2 - (-2)$, which is $4 + 2$, which is $\geq 0$. But might be that I must be wrong somewhere. Please correct me. Thanks.

share|cite|improve this question
Yes. The answer is just $[0, 1]$. – Sangchul Lee May 17 '14 at 18:52
@sos440 Thanks! I had been stuck on that for so long. – Gaurang Tandon May 17 '14 at 18:54
up vote 1 down vote accepted

$w^2-w\le 0$

$w(w-1)\le 0$

$0\le w\le 1$

The answer given is wrong.

share|cite|improve this answer

In general, when you want to solve an inequality $p(x) \leq 0$ where $p$ is a polynomial, you first solve $p(x) = 0$ for the roots $x$ and then see if $p$ is positive or negative on each side of the roots. For $x^2 - x$, the roots are $x = 0,1$. It is then easy to check that $x^2 - x > 0$ when $x < 0$ or when $x > 1$, and $x^2 - x < 0$ when $0 < x < 1$. So the solution is just $0 \leq x \leq 1$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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