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?


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  • $\begingroup$ When the equation has more than two roots, it means that it intercepts the $x$-axis more than two times. But assuming that none of the coefficients are zero, you can prove that it can intercept the $x$-axis (or any other horizontal line, in fact) only twice or not at all. Thus the coefficients must all be zero. $\endgroup$ – shardulc May 22 '15 at 18:35
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    $\begingroup$ Welcome to StackExchange! What have you tried to do? What have you found out about the problem? Try to give us as much info in the future so we can get in your mindset. Anyways, I would recommend looking up how many roots can a parabola have and properties of parabolas to get a geometric intuition based on the quadratic formula. Also, the looking up the fundemental theorem of algebra and noticing it only applies towards nonzero polynomials may be fruitful. $\endgroup$ – user2154420 May 22 '15 at 18:39
  • $\begingroup$ An equation $A=B$ is only said to be identity if the solution set is the domain of the equation. :) $\endgroup$ – Sufyan Naeem May 22 '15 at 18:55

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 $x$.

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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    $\begingroup$ It is not said anywhere that all multiplicities should be 1. $\endgroup$ – Celsius May 22 '15 at 18:37

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.


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 $a=b=c=0$.

  • $\begingroup$ The proof with the Vandermonde matrix is also elementary, and very convenient - see here. Also the factor theorem suffices. $\endgroup$ – Dietrich Burde May 22 '15 at 18:58
  • $\begingroup$ @DietrichBurde The OP tagged this as algebra-precalculus, which makes me think that matrices in general may not be part of their mathematical vocabulary. $\endgroup$ – user90667 May 22 '15 at 19:02
  • $\begingroup$ OK, then no matrices. In this case, you only need induction, see the answer of Dan Petersen here. $\endgroup$ – Dietrich Burde May 22 '15 at 19:04

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