# Using Rolle's theorem prove that if $b^2<3ac$, then there is exactly one root to $f(x) \equiv ax^3+bx^2+cx+d=0$

Using Rolle's theorem prove that if $$b^2<3ac$$, then there is exactly one root to $$f(x) \equiv ax^3+bx^2+cx+d=0$$

I literally have no idea how to use Rolle's Theorem.

Rolle's Theorem: if the function $$f \in C[a,b]$$ and differentiable on $$(a,b)$$, and $$f(a)=f(b)$$, then there exists $$c \in (a,b)$$ for which $$f'(c)=0$$.

What I did so far:

$$f'(x)=3ax^2+2bx+c$$,

$$D=4b^2-12ac>4b^2-4b^2=0 \implies f'(x)=0$$ has exactly two roots.

If I use given inequality, I don't prove anything what I need. Any help would be appreciated.

I have looked at a similar example Prove that if ab > 0 then the equation $ax^3 + bx + c = 0$ has exactly one root by Rolle's theorem but I didn't find it a lot helpful.

If $$b^2 <3ac$$,then your derivative $$f'(x)=3ax^2+2bx+c$$ does not have a real root.

On the other hand if $$f(x)=0$$ had more that one real solutions, according to the Roll's theorem its derivative had to have at least one real root between those two. Therefor $$f(x)$$ has only one real root.

Note that a third degree polynomial always have at least one real root because of its end behavior.

The hypothesis implies that $$a \neq 0$$. If $$a >0$$ then $$f(x) \to \infty$$ as $$x \to \infty$$ and $$f(x) \to -\infty$$ as $$x \to -\infty$$ so there is at least one root. Similar argument works for $$a <0$$.

If there are two roots then $$f'(x)=0$$ for some $$x$$. But the quadratic $$3ax^{2}+2bx+c=0$$ does not have any real roots since $$4b^{2}-12ac=4(b^{2}-3ac)<0$$. This completes the proof.

Oh, there is a mistake:

if $$b^2 - 3ac < 0$$ (the same as your inequality), then $$D=4b^2 - 12ac = 4(b^2 - 3ac) < 0$$, so the derivative has no real roots.