If a quadratic equation $ax^2+bx+c=0$ has more than two roots, then $a=b=c=0$

Question

If a quadratic equation $ax^2+bx+c=0$ has more than two roots, then it is an identity i.e. it is true for all values of $x$ and $a=b=c=0$.

What is a proof of this?

Answer 1

Let the three roots be $x_1,x_2,x_3$.

Method $1$: Let $f(x) = ax^2+bx+c$. Since $x_1$ and $x_2$ are roots, this means $f(x) = (x-x_1)(x-x_2)g(x)$. Since $f(x)$ has degree $2$, this forces $g(x)$ to be a constant say $k$. Further, we have $f(x_3) = 0$. This means $k(x_3-x_1)(x_3-x_2) = 0$. Since $x_3 \neq x_1$ and $x_3 \neq x_2$, this forces $k$ to be zero. Hence, $f(x) \equiv 0$ for all $x$.

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Method $2$ Here we shall assume that $x_i$'s are distinct. This means we have \begin{align} ax_1^2 + bx_1 + c & = 0\\ ax_2^2 + bx_2 + c & = 0\\ ax_3^2 + bx_3 + c & = 0 \end{align} where $x_i$'s are distinct. We now have a linear system for $a,b,c$ with the right hand side being zero. Writing it in matrix form $$ \begin{bmatrix} x_1^2 & x_1 & 1\\ x_2^2 & x_2 & 1\\ x_3^2 & x_3 & 1 \end{bmatrix} \begin{bmatrix} a\\ b\\ c \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} $$ The determinant of $\begin{bmatrix} x_1^2 & x_1 & 1\\ x_2^2 & x_2 & 1\\ x_3^2 & x_3 & 1 \end{bmatrix}$ is $(x_1-x_2)(x_2-x_3)(x_3-x_1)$, which is non-zero since $x_i$'s are distinct. This means the only solution for $\begin{bmatrix} a\\ b\\ c \end{bmatrix}$ is $$\begin{bmatrix} a\\ b\\ c \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} $$

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Answer 2

This is not true in general. The polynomial $x^2+1$ has more than two zeroes over the quaternions $\mathbb{H}$, but is not identical zero. I suppose you assume implicitly that the domain is a field ? Over a field every nonzero polynomial of degree $n$ has at most $n$ zeroes, see here. The proof uses the Vandermonde matrix. Hence if a quadratic polynomial has more than two roots it must be identically zero.

Answer 3

Here's a more elementary proof. It's more cumbersome than other answers here, but it gets the job done...

Assume there are three distinct roots of your quadratic $r_1, r_2, r_3$. Then we have \begin{align*} ar_1^2 + br_1 + c &= 0\\ ar_2^2 + br_2 + c &= 0\\ ar_3^2 + br_3 + c &= 0 \end{align*} Subtracting the second equation from the first, we get $a(r_1^2 - r_2^2) + b(r_1-r_2) = 0$, which implies $(r_1-r-2)[a(r_1+r_2)+b] = 0$. Doing the same thing with the first and third equation, we get the system \begin{align*} (r_1-r-2)[a(r_1+r_2)+b] &= 0 \\ (r_1-r-3)[a(r_1+r_3)+b] &= 0 \end{align*} Since $r_1,r_2,r_3$ are unequal, this implies \begin{align*} a(r_1+r_2)+b &= 0 \\ a(r_1+r_3)+b &= 0 \end{align*} Subtracting the first equation from the second, we get $a(r_1-r_3) = 0$. Since $r_1 \not= r_3$, we have $a=0$. Which is huge. Plugging this back into the system at the top, we have \begin{align*} br_1 + c &= 0\\ br_2 + c &= 0\\ br_3 + c &= 0 \end{align*} If we had $b\not= 0$, this would imply $\frac{-c}{b} = r_1 = r_2 = r_3$, which is a clear contradiction. Therefore, $b = 0$, and we are left with $c=0$. Therefore, we have shown $a=b=c=0$.