Using Rolle's theorem and IVT, show that $x^4-7x^3+9=0$ has exactly $2$ roots. Using Rolle's Theorem and the Intermediate Value Theorem, show that $x^4-7x^3+9=0$ has exactly two roots.


I know how to prove that this equation has at least two real roots by using IVT, but my problem is how do I use Rolle's theorem to prove that it has at most two real roots? I've used Rolle's theorem to prove a function has one real root, but how do I do it with more than one root.
Set $f(x)=x^4-7x^3+9$. The derivative 
$$f'(x)=4x^3-21x^2=x^2(4x-21)$$
has the sign of $4x-21$, so $f(x)$ is decreasing up to $\frac{21}{4}$, attains a unique minimum $f\bigl(\frac{21}4\bigr)$ and is finally increasing. By the I.V.T. and the definition of the monotony of a function, the equation has exactly two roots if this minimum is negative, none if it is positive and a double root if the minimum is $0$.

This can be generalised to any quartic polynomial $ax^4+bx^3+c$.
The derivative is $4x^3-21x^2=x^2(4x-21)$ is negative if $x&lt; 21/4$ and positive if $x&gt;21/4$, $f(21/4)&lt;0$ and $lim_{-\infty}f(x)=+\infty$ $lim_{+\infty}f(x)=+\infty$. We deduce that $f$ is strictly decreasing in $(-\infty,21/4)$ and has one root in this interval, $f$ is strictly increasing in $(21/4,+\infty)$  and has one root in this interval.