# Prove that if ab > 0 then the equation $ax^3 + bx + c = 0$ has exactly one root by Rolle's theorem

I have deduced a proof as stated below and am not sure if it is correct, therefore need some advice.

Proof: Let $$f(x) = ax^3+bx+c = 0$$. $$f(x)$$ is continuous since it is a polynomial and it is differentiable since it has a limit. Assume $$f(x)$$ has $$2$$ roots, $$f(a) = 0$$ and $$f(b) = 0$$, there is a point $$d\in(a,b)$$ such that $$f'(d) = 0$$.

$$f'(x) = 3ax^2+b$$ Since $$ab>0$$, $$a$$ and $$b$$ must be both positive or both negative. $$f'(d) = 3a(d)^2+b = 3ad^2+b$$ not equal to $$0$$ instead $$>0$$ since any values of $$d$$ for $$d^2$$ will be positive.

Likewise $$f'(d) = 3a(d)^2 + b$$ will not equal to $$0$$ instead $$<0$$ for all negative values of $$a$$ and $$b$$.

Hence, a contradiction, $$f$$ has exactly one root.

1. You should begin the proof with "Let $$f(x)=ax^3+bx+c$$". It is commonly understood that any reader should understand that this means "Let $$f(x)=ax^3+bx+c$$ for all $$x$$ (i.e.for all $$x\in \Bbb R$$ )"...rather than writing "$$f(x)=ax^3+bx+c=0$$".

2. You should not use $$a,b$$ for possible values of $$x$$ for which $$f(x)=0$$ because the letters $$a,b$$ are already being used for something else. "Assume $$f(x_1)=f(x_2)=0$$ with $$x_1 Then by Rolle's Theorem there exists $$d\in (x_1,x_2)$$ such that $$f'(d)=0$$."

3. Allowing for the poor grammar in the rest of the proof, it is correct EXCEPT for the last line. You have shown that there is at MOST one $$x$$ such that $$f(x)=0.$$ But you have not shown that there is also at LEAST one $$x$$ such that $$f(x)=0.$$

• Thanks noted on that. – Alexander Oct 15 '18 at 7:45
• This is a good answer addressing OP's question instead of solving OP's problem. – Surb Feb 21 '19 at 0:14

The inequality after first derivative can be proven with discriminant:

\begin{align*} f'(x) &= 3ax^2 + b\\ D &= 0-4\cdot(3a)\cdot b\\ &< 0 \end{align*}

So $$f'(x) \ne 0$$ for all $$x$$.

• Does the OP prove that $f$ has any real roots? – Michael Burr Oct 14 '18 at 10:22
• @MichaelBurr. No. He did not. I pointed that out in my answer. – DanielWainfleet Oct 14 '18 at 10:33
• Thanks noted on that. – Alexander Oct 15 '18 at 7:45