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

Question

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 $ab < 0.$

Answer 1

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?

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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.