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Prove that if $ab < 0$ then the equation $ax^{3} + bx + c = 0$ has at most three real roots.

I would need verification on the proof below, thanks!

Proof:

Let $f(x) = ax^{3} + bx + c.$

Assume that $f(x)$ has $4$ distinct roots, $f(p) = f(q) = f(r) = f(s) = 0$, there is a point $x_1$ element of $(p,q)$ such that $f'(x_1) = 0; x_2$ element of $(q, r)$ such that $f'(x_2) = 0$; $x_3$ element of $(r,s)$ such that $f'(x_3) = 0.$

Since $ab < 0$ then there are two possibilities where $a>0$ and $b<0$ or $a<0$, $b>0.$

$$f'(x) = 3ax^{2}+b.$$

If $|3ax^{2}|= |b|$ where $3ax^{2} > 0$ and $b < 0$, then $f'(x) = 0.$

If $|3ax^2|= |b|$ where $3ax^{2} < 0$ and $b > 0$, then $f'(x) = 0.$

This is not true because the equation $f'(x) = 0$ has only two roots.

Hence the given equation has at most three real roots when $ab < 0.$

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    $\begingroup$ How can it have more than 3 roots? $\endgroup$ – Wuestenfux Oct 15 '18 at 7:47
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    $\begingroup$ A degree 3 polynomial can't have more than 3 roots, no matter what the coefficients are. $\endgroup$ – Gerry Myerson Oct 15 '18 at 7:47
  • $\begingroup$ Please advice how can this be proved. Thks $\endgroup$ – Alexis Oct 15 '18 at 8:08
  • $\begingroup$ If $r$ is a root, then $x-r$ is a factor. A polynomial of degree 3 can't have more than 3 factors. Polynomials over a field form a unique factorization domain. $\endgroup$ – Gerry Myerson Oct 15 '18 at 11:43
  • $\begingroup$ I wouldn't downvote this question myself, but I think that I can understand why others have---it seems to be missing some context. The question doesn't make much sense if you can assume the fundamental theorem of algebra, so it might be worth noting that the FTA should not be used; the proof seems to be making an argument via the derivative, so I suspect that this is an MVT problem from a calc class? That, also, would be nice context to have. In general, when you are asking a question, you should be clear about what tools you are expected to use in order to answer that question. $\endgroup$ – Xander Henderson Oct 15 '18 at 13:55
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First, a minor bit of pedantry. You state the result as

Prove that if $ab < 0$ then the equation $ax^{3} + bx + c = 0$ has at most three real roots.

While I understand your meaning, it is not generally correct to say that an equation has a root. You can either say that the function $f$, defined by the formula $f(x) = ax^3 + bx + c$ has a root; or that the equation $ax^3 + bx + c = 0$ has a solution. That being said, you may have copied this directly from a book, and this might be considered a matter of style more than a matter of mathematics.

Also, I don't see what the condition $ab < 0$ tells us in this problem. I would leave it out, and prove the stronger result, i.e. the equation has at most three real roots for any choice of $a$, $b$, and $c$, just so long as at least one of those constants is nonzero.

As to your argument, you start your proof by stating:

Assume that $f(x)$ has $4$ distinct roots, $f(p) = f(q) = f(r) = f(s) = 0$, there is a point $x_1$ element of $(p,q)$ such that $f'(x_1) = 0; x_2$ element of $(q, r)$ such that $f'(x_2) = 0$; $x_3$ element of $(r,s)$ such that $f'(x_3) = 0.$

In this step, you should perhaps indicate that $p < q < r < s$. You seem to make use of this fact when you find zeroes of $f'$ in the intervals $(p,q)$, $(q,r)$, and $(r,s)$. If you don't assume that the endpoints are ordered as I have done, then you cannot guarantee that the three zeroes you find are unique.

As a matter of taste, I would also more explicitly state that you are applying Rolle's theorem.

Since $ab < 0$ then there are two possibilities where $a>0$ and $b<0$ or $a<0$, $b>0.$ $$f'(x) = 3ax^{2}+b.$$

If $|3ax^{2}|= |b|$ where $3ax^{2} > 0$ and $b < 0$, then $f'(x) = 0.$

If $|3ax^2|= |b|$ where $3ax^{2} < 0$ and $b > 0$, then $f'(x) = 0.$

This is not true because the equation $f'(x) = 0$ has only two roots.

I think that you are moving towards the correct argument, namely that a quadratic polynomial can have at most two real roots, but you have found three via Rolle's theorem, which is a contradiction. That being said, I don't follow this part of the argument. Are you claiming that $f'(x) = 0$ for all $x$? Or are you trying to say something else?


For my taste, I would probably state your argument as follows:

Prove that if $a$, $b$, and $c$ are not all zero, then the equation $ax^{3} + bx + c = 0$ has at most three real roots.

Proof: First, dealing with the trivial cases, note that if $a = 0$, then either

  • $f(x) = bx + c$, with $b \ne 0$, which has one real root (namely, $x = -c/b$), or
  • $f(x) = c$, with $c \ne 0$, which has no real roots.

Hence if $a = 0$, then $f$ has fewer than three real roots.

So suppose that $a \ne 0$. Then $f'(x) = 3ax^2 + b$, which has at most two real roots: $$x = \pm\sqrt{\frac{b}{3a}}.$$ (Note: if $ab < 0$, then $f'$ will have exactly two real roots; otherwise, it will have no real roots. In either case, it has at most two real roots. In short, the original assumption that $ab < 0$ ensures that the first derivative has two real roots.)

Assume for contradiction that $f$ has at least four real roots: say $p < q < r < s$. By Rolle's theorem, there exist distinct real values $$ x_1 \in (p,q), \qquad x_2 \in (q,r), \qquad\text{and}\qquad x_3 \in (r,s) $$ such that $f'(x_i) = 0$. But then $f'$ has three real roots, which is a contradiction, as $f'$ can have at most two real roots.

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  • $\begingroup$ Thank you for the comments given. Can i verify if it is still necessary to write the "Assume for contradiction that f has at least four real roots: say p<q<r<s. By Rolle's theorem, there exist distinct real values ......" part? $\endgroup$ – Alexis Oct 16 '18 at 12:30
  • $\begingroup$ What do you mean by "necessary"? The goal of a proof is to communicate an idea. In writing that first sentence, I am communicating a lot of ideas in a small amount of text: the proof is going to go by contradiction; the contradiction will come from the fact that $p$, $q$, $r$, and $s$ are assumed to be roots; I have (without loss of generality) assumed that the roots are ordered in a particular way. It may not be "necessary" to do this, but it helps to communicate to the reader what is going on. Ditto the line about Rolle's theorem. $\endgroup$ – Xander Henderson Oct 16 '18 at 15:58
  • $\begingroup$ sure thanks alot. $\endgroup$ – Alexis Oct 19 '18 at 15:05

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