Suppose that $P: \mathbb{R} \rightarrow \mathbb{R}$ is a real polynomial of degree exactly $2$. Prove $P$ has at most two roots.

Let $P(x)=a_2 x^2 +a_1 x +a_0$ for all $x \in \mathbb{R}$. I tried to assume there are more than $2$ roots and contradict it using Rolle's theorem but it wasn't working out. I am stuck.

  • 1
    $\begingroup$ Suppose by contradiction that it has three roots. Then... $\endgroup$ – abiessu May 26 '15 at 23:29
  • $\begingroup$ Or you can always break out the hammer and use the fundamental theorem of algebra, but I seriously doubt your teacher intends you to do that :) $\endgroup$ – Alan May 26 '15 at 23:33
  • $\begingroup$ ...the derivative $P'$ which is a polynomial of degree $1$ has two distinct roots. $\endgroup$ – user226387 May 26 '15 at 23:34
  • $\begingroup$ @math what does that mean though? I don't see how P' shows anything about the roots... $\endgroup$ – snowman May 26 '15 at 23:37
  • $\begingroup$ My comment completes that of abiessu. $\endgroup$ – user226387 May 26 '15 at 23:39

I was going to go with the systems of equations approach, but got beaten to the chase. So here is a more "calculus" type answer.

Let the polynomial be $P(x)=ax^2+bx+c$. Then we can write the derivative as $P'(x)=2ax+b$.

Without loss of generality suppose that $a>0$. If it isn't then just multiply the polynomial by $-1$, which will not change its roots.

Note that $P'(x) > 0 $ when $x>-b/2a$ and $P'(x)<0$ when $x<-b/2a$. This means that $P$ is strictly increasing on $(-\infty,-b/2a)$ and then strictly decreasing on $(-b/2a,\infty)$. From this we should be able to conclude that it can only cross the $x-axis$ at most twice, but that is not yet a proof.

With the above in mind suppose that there are three roots $x_1,x_2$, and $x_3$. Then Rolle's theorem tells us that the derivative of $P$ is zero in the intervals $[x_1,x_2]$ and $[x_2,x_3]$. This is impossible because the derivative is linear and only has one root. Therefore there cannot be three roots.

  • $\begingroup$ I think this is the intended approach in a calculus class. +1. (Except you should have $2ax+b$, of course.) $\endgroup$ – Ian May 26 '15 at 23:38

Suppose there were three (distinct) roots were $a,b,c$. Then

$$\begin{pmatrix} a^2 & a & 1 \\ b^2 & b & 1 \\ c^2 & c & 1 \end{pmatrix}\begin{pmatrix} a_2 \\ a_1 \\ a_0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$

As $\begin{pmatrix} a_2 \\ a_1 \\ a_0 \end{pmatrix}$ is a non-zero vector, that must mean the the matrix $$M = \begin{pmatrix} a^2 & a & 1 \\ b^2 & b & 1 \\ c^2 & c & 1 \end{pmatrix}$$ has non-trivial kernel or equivalently it has zero determinant.

However $\det M = -(a-b)(b-c)(c-a) \neq 0$.

  • $\begingroup$ I find it amazing that this was one a real analysis past paper... $\endgroup$ – snowman May 26 '15 at 23:32
  • 1
    $\begingroup$ Lots of ways to skin this cat. $\endgroup$ – Simon S May 26 '15 at 23:32


This is purely algebraic:

$\alpha$ is a root of $p(x)$ if and only if $p(x)$ is divisible by $x-\alpha$. Then remember $\deg(p(x)q(x))=\deg p(x)+\deg q(x)$.

Edit: This result can be generalised further: a polynomial over any field (or integral domain) of degree $d$ has at most $d$ roots.


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.