# Polynomial of degree $2$ has at most $2$ roots

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.

• Suppose by contradiction that it has three roots. Then... – abiessu May 26 '15 at 23:29
• 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 :) – Alan May 26 '15 at 23:33
• ...the derivative $P'$ which is a polynomial of degree $1$ has two distinct roots. – user226387 May 26 '15 at 23:34
• @math what does that mean though? I don't see how P' shows anything about the roots... – snowman May 26 '15 at 23:37
• My comment completes that of abiessu. – 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.

• I think this is the intended approach in a calculus class. +1. (Except you should have $2ax+b$, of course.) – 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 $\det M = -(a-b)(b-c)(c-a) \neq 0$.

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

Hint:

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.