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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.

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3 Answers 3

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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.

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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.

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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.

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