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Cubic depressed form with equation $f(x) = x^3 + px + q$

The question is, when $p$ is positive, will the function have $3$ real roots ? or does it have to have $1$ real and $2$ complex roots?

My question is not the same as Maths cubic equation discriminant....

I need to know the possibilities of roots when p is positive.

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Tommy Lassa
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1 Answers1

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We have $f'(x) = 3x^2+p \ge p > 0 $. Consequently, $f$ is strictly increasing.

  • $f(x)=0$ has at most one real solution. Otherwise take two distinct real solution $x_1<x_2$. Since $f$ is strictly increasing we have $0=f(x_1)<f(x_2)=0$. Absurd.
  • Since $$\lim_{x\to +\infty}f(x) = +\infty,\quad\text{and }\quad \lim_{x\to -\infty}f(x) = -\infty.$$ The function $f$ is continuous and $f(x)$ goes from $-\infty$ to $+\infty$ when $x$ goes from $-\infty$ to $+\infty$. By continuity, $f(x)$ has to take the value $0$ at least once.

Finally, there is exactly one solution to the equation $f(x)=0$ for $x\in \mathbb{R}$.

user37238
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  • Yes, but that is only for the extrema. If p is positive, there will not be any min or max so the function can only have 1 real 2 complex or 1 triple roots? Is that correct? Did I miss any possibilities? – Tommy Lassa Jul 28 '15 at 13:27
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    Your question is about positive $p$. – user37238 Jul 28 '15 at 13:29
  • sorry typo. i am confused. Please look at this page http://math.stackexchange.com/questions/1371510/maths-cubic-equation-discriminant . It says p has to be negative in order to let 2 extrema to occur. So I am not sure... – Tommy Lassa Jul 28 '15 at 13:33
  • @TommyLassa If there is something you don't understand there, you may ask for clarification there. Or you can ask a different question about this specific point. – user37238 Jul 28 '15 at 13:35
  • Yes, but it is a different question. I want to know the different possibilities of cubic roots according to different value of p. My previous post merely explain the origin of discriminant. – Tommy Lassa Jul 28 '15 at 13:37
  • Now that I've developed my answer, you understand why in the case where $p>0$, the function $f$ has $1$ real root and $2$ complex roots? – user37238 Jul 28 '15 at 13:42
  • sorry I dont understand anything you write :( sorry. May you guide me a little bit? – Tommy Lassa Jul 28 '15 at 13:55
  • You don't understand anything I write? I find it hard to believe. Please tell me precisely what you don't understand. – user37238 Jul 28 '15 at 14:18
  • wait I understand :). so when p is negative, cubic can have multiple roots? – Tommy Lassa Jul 28 '15 at 14:34
  • What about $f(x)=x^3-x$ for example? Or $f(x)=x^3$? – user37238 Jul 28 '15 at 14:36
  • so they are some special cases? what is special about x^3-x? p is -1 so we know it has to have 3 real roots. But when p and q are zero there will be 1 triple roots ? – Tommy Lassa Jul 28 '15 at 14:41