How can we prove that a quadratic equation has at most 2 roots?

Question

A quad equation can be factored into two factors containing $x $, but how can we prove that there no other sets of different factors yielding OTHER VALUES OF $X $?

Answer 1

Any quadratic equation has at most 2 roots because any polynomial equation $f(x)$ ($f(x)$ defines the polynomial) will have $\deg f(x)$ roots to it. Because of that, the degree of quadratics is 2, so there are a maximum of 2 roots in $\mathbb{C}$. To prove this, let $\alpha$, $\beta$, and $\gamma$ represent three roots of any quadratic equation in the form $ax^2 + bx + c = 0$. So, each $\alpha$, $\beta$, and $\gamma$ will satisfy said quadratic equation: $$\begin{align} a\alpha^2 + b\alpha + c & = 0 \tag 1 \label{1}\\ a\beta^2 + b\beta + c & = 0 \tag 2 \label{2} \\ a\gamma^2 + b\gamma + c & = 0 \tag 3 \label{3} \end{align} $$ If you subtract equation $\ref{2}$ from $\ref{1}$, you get $$\begin{align} a\alpha^2 + b\alpha + c - (a\beta^2 + b\beta + c) & = 0 \\ \implies a(\alpha^2 - \beta^2) + b(\alpha - \beta) & = 0 \\ a(\alpha - \beta)(\alpha + \beta) + b(\alpha - \beta) & = 0 \\ (\alpha - \beta)(a(\alpha + \beta) + b) & = 0 \\ a(\alpha + \beta) + b & = 0 && (\alpha - \beta \neq 0) \tag 4 \label{4} \end{align} $$ If you subtract equation $\ref{3}$ from $\ref{2}$, you get $$\begin{align} a\beta^2 + b\beta + c - (a\gamma^2 + b\gamma + c) & = 0 \\ \implies a(\beta^2 - \gamma^2) + b(\beta - \gamma) & = 0 \\ (\beta - \gamma)(a(\beta + \gamma) + b) & = 0 \\ a(\beta + \gamma) + b & = 0 && (\beta - \gamma \neq 0) \tag 5 \label{5} \end{align} $$ Now, when you subtract equation $\ref{5}$ from $\ref{4}$, you get $$\begin{align} a(\alpha - \gamma) & = 0 \\ \implies \alpha = \gamma \end{align} $$ This is an impossible situation because $\alpha$ and $\gamma$ are both distinct roots of the quadratic equation and $a \neq 0$. Therefore, there is a maximum of 2 solutions. Quod erat demonstrandum.

References:

• My question related to this: Why does the equation $\frac{-x^2 + 2x}{5x - 4} = 6$ have 2 solutions?

• Show that Quadratic Equations cannot have more than 2 Roots- AskMath

Answer 2

This question is equivalent to proving the uniqueness of the factorization of a polynomial over a field (which for simplicity's sake I'm going to assume is $\Bbb C $ since you haven't specified - which avoids irreducible polynomials).

So assume there exists $x - r_1$, $x - r_2$ both divide $ax^2 + bx + c $ with some remainder $q(x)$, $p(x)$ respectively.

$(x - r_1)*q(x) = ax^2 + bx + c$

$(x - r_2)*p(x) = ax^2 + bx + c$

So $(x - r_1)*q(x) = (x - r_2)*p(x)$

Then it follows that say $(x - r_1) | (x - r_2)*p(x)$

Can you complete the proof from here?