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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.

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3  
Yes. The answer is just $[0, 1]$. –  sos440 May 17 at 18:52
    
@sos440 Thanks! I had been stuck on that for so long. –  Gaurang Tandon May 17 at 18:54

2 Answers 2

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.

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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 $0 \leq x \leq 1$.

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