1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.