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 $X$?

• If a non constant polynomial $p$ has a root $r$ then we can write $p(x) = (x-r)q(x)$ for some polynomial $q$ of degree less than the degree of $p$. Hence there can be at most two factors. – copper.hat Mar 8 '16 at 16:55
• This is a proof that doesn't rely on any theorem: assume $(x-a)(x-b)=(x-a')(x-b')$. Then putting $x=0$ you have $ab=a'b'$ and putting $x=1$ you have $a+b=a'+b'$. This implies that $(a-b) ^2=(a+b)^2-4ab=(a'-b')^2$. From $a+b=a'+b'$ and $a-b=\pm(a'-b')$ you can derive $\{a,b\}=\{a',b'\}$. – Marco Disce Mar 9 '16 at 15:21
• @Marco Disce Nice – N.S.JOHN Mar 9 '16 at 16:20
• if there were more than 2 roots, you would get $f(x)=(x-r1)(x-r2)(x-r3)$, this when expanded results in a polynomial of 3rd degree if the value r3 is different from r2 and r1. – NoChance Apr 16 '16 at 22:35
• @NoChance That is not the essence of my question. I was asking why $(x-r1)(x-r2)\neq (x-r3)(x-r4)$ – N.S.JOHN Apr 17 '16 at 4:57

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:

• Quite the circular argument, isn't it? – Deathkamp Drone Apr 16 '16 at 17:14
• What do you mean? – Obinna Nwakwue Apr 16 '16 at 18:17
• To say that it has at most 2 roots because it has degree 2 is the same thing as saying that it has at most 2 roots because it is a quadratic equation. But the question is about why is that so. – Deathkamp Drone Apr 16 '16 at 19:01
• Edited, and I included a proof. – Obinna Nwakwue Apr 16 '16 at 22:31

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?

• I was just thinking to myself that the present comment and answer don't really get at the essence of what the OP was asking when your answer popped up. – Dave L. Renfro Mar 8 '16 at 17:24
• @DaveL.Renfro is right. By the way, I don't understand the last part. Can you elaborate on that. Also, take a look at my proof. I think the OP would better understand mine. Also, you're right; when we are working with fields of quadratics, we would most likely be working in the field of $\mathbb{C}$. – Obinna Nwakwue Apr 16 '16 at 22:44

Suppose $\,f(x)\,$ is a polynomial of degree $2$ with coefficients in a field $\,F$ (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 field). So $\,f\,$ has at most $2$ distinct roots.

Bifactor Theorem $\$ Suppose that $\rm\,a,b\,$ are elements of a field $\rm\,F\,$ 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 $\rm\: \color{#C00}{a\!-\!b\ne 0},$

$$\begin{eqnarray}\rm\:f(b)= 0 &\ \Rightarrow\ &\rm f(x)\, =\, (x\!-\!b)\,g(x)\ \ for\ \ some\ \ g\in F[x]\\ \rm f(a) = (\color{#C00}{a\!-\!b})\,g(a) = 0 &\Rightarrow&\rm g(a)\, =\, 0\,\ \Rightarrow\,\ g(x) \,=\, (x\!-\!a)\,h(x)\ \ for\ \ some\ \ h\in F[x]\\ &\Rightarrow&\rm f(x)\, =\, (x\!-\!b)\,g(x) \,=\, (x\!-\!b)(x\!-\!a)\,h(x)\end{eqnarray}$$