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

• 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? May 8, 2019 at 7:35
• 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. 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.