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.


4 Answers 4


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 $ax^4+bx^3+c$.

  • $\begingroup$ I see how you have shown that there are at least 2 roots, but how do you show that there cannot be 3 or more roots? $\endgroup$
    – makansij
    May 8, 2019 at 7:35
  • $\begingroup$ That is because a function which is monotonic on an intaval cannot have more one root on this interval. This is geometrically obvious, and is proved using Rolle's theorem: if there wer a second root, there oumd also be a root of the derivative between the roots. $\endgroup$
    – Bernard
    May 8, 2019 at 9:29

Between two roots of a polynomial the derivative must vanish at least once, by Rolle’s theorem.

Since the derivative vanishes at $0$ and $21/4$, because $f'(x)=x^2(4x-21)$, there can be as much as three roots.

A refined version of Rolle’s theorem, however, tells us that the derivative must vanish at a local maximum or minimum between two roots; thus we can exclude $0$, where the derivative doesn't change sign (the function is decreasing in a neighborhood of $0$). Hence the roots are at most two.

Their number is even, because clearly the function hasn't multiple roots; so we have either no root or two roots.

Since $$ f(0)=9, \qquad f(6)=-207, \qquad f(7)=9 $$ the IVT tells you that there is a root in $(0,6)$ and a root in $(6,7)$.


Let $f(x)=x^4-7x^3+9$

$D=\mathbb{R}$ ( $f(x)$ is continuos on $\mathbb{R}$)

//Prove $f(x)$ has at least 2 real roots//

We have:

$f(1) = 3 >0$; $f(2) = -31 <0$

$\therefore$ $f(x)$ has at least one real root between $1$ and $2$ (intermediate value theorem) $(1)$

$f(6) = -207 <0$; $f(7)=9>0$

$\therefore$ $f(x)$ has at least one real root between $6$ and $7$ (intermediate value theorem) $(2)$

From $(1)$ and $(2)$ ==> $f(x)$ has at least 2 real roots.

//Prove $f(x)$ has exactly 2 real roots//

Consider $f'(x)=0$ ==> $4x^3-21x^2=0$

<=> $x=0$ $\cup$ $x=\frac{21}{4}$

We have $f(0)=9 >0$ and $f(\frac{21}{4})=-\frac{62523}{256} <0$

==> $f(x)$ cannot has more than $3$ real roots.

However, the second root we've proved earlier (between $6$ and $7$) > $\frac{21}{4}$ (the graph of $f(x)$ is now increasing to infinity)

==> $f(x)$ has exactly 2 real roots ==> QED.

(I recommend you draw a graph on GeoGebra or Desmos to understand clearly)


The derivative is $4x^3-21x^2=x^2(4x-21)$ is negative if $x< 21/4$ and positive if $x>21/4$, $f(21/4)<0$ 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.


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.