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


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*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
A: Let $\,f(x)\,$ is a polynomial of degree $2$ with coefficients in a field or domain $\,F$ (e.g. $\Bbb Q,\Bbb R,\Bbb C)$ and suppose that $\,f\,$ has $\,2\,$ distinct roots $\,a,b.\,$ By the Bifactor Theorem below we deduce that $\,f(x) = c(x\!-\!a)(x\!-\!b)\,$ for $\,c\in F.\,$ Thus if $\,d\neq a,b\,$ then $\,f(d) = c(d\!-\!a)(d\!-\!b)\ne 0\,$ since each factor is $\ne 0\,$ (recall $\,x,y\ne 0\,\Rightarrow\,xy\ne 0\,$ in a domain). So $\,f\,$ has at most $2$ distinct roots.
Bifactor Theorem $\ $ Suppose that $\rm\,a,b\,$ are elements of a domian $\rm\,F\,$ and $\rm\:f\in F[x],\,$ i.e. $\rm\,f\,$ is a polynomial with coefficients in $\rm\,F.\,$ If $\rm\ \color{#C00}{a\ne b}\ $ are elements of $\rm\,F\,$ then
$$\rm f(a) = 0 = f(b)\ \iff\ f\, =\, (x\!-\!a)(x\!-\!b)\ h\ \ for\ \ some\ \ h\in F[x]$$
Proof $\,\ (\Leftarrow)\,$ clear. $\ (\Rightarrow)\ $ Applying  Factor Theorem twice, while canceling $\rm\: \color{#C00}{a\!-\!b\ne 0},$
$$\begin{eqnarray}\rm\:f(b)= 0 &\ \Rightarrow\ &\rm f(x)\, =\, (x\!-\!b)\,g(x)\ \ for\ \ some\ \ g\in F[x]\\
\rm f(a) = (\color{#C00}{a\!-\!b})\,g(a) = 0 &\Rightarrow&\rm g(a)\, =\, 0\,\ \Rightarrow\,\ g(x) \,=\, (x\!-\!a)\,h(x)\ \ for\ \ some\ \ h\in F[x]\\
&\Rightarrow&\rm f(x)\, =\, (x\!-\!b)\,g(x) \,=\, (x\!-\!b)(x\!-\!a)\,h(x)\end{eqnarray}$$
