Why can a quadratic equation have only 2 roots? It is commonly known that the quadratic equation $ax^2+bx+c=0$ has two solutions given by: $$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$$ But how do I prove that another root couldn't exist?
I think derivation of quadratic formula is not enough....
 A: 
I think derivation of quadratic formula is not enough....

Yes it is. The derivation is of the form if $ax^2+bx+c=0$, then $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$. The derivation is a proof if you pay attention.
The trickiest step is simply that if $y^2 = k$ for $k \geq 0$ then $y = \pm \sqrt k$, if you do not take this as evident.
A: I think I can simplify the 'polynomial long division' answer.  The special case of polynomial long division says that, for any polynomial $P$, and any real number $a$,
$$P(x) = Q(x)(x-a) + R$$
for some polynomial $Q$ and constant $R$ (use long division to divide $P$ by $x-a$ and observer that $R$ is required to be a constant since it must have lower degree than $x-a$, which is a first-degree polynomial).
The above is true for all $x$, so substituting $x=a$ we get
$$P(a) = Q(a)(a-a) + R$$
Obviously, $a-a=0$, so $R=P(a)$.
If $a$ is a 0 of $P$ ($P(a)=0$), then $R=0$, so $x-a$ divides $P(x)$.
Now, if we have any distinct root $a$ of a quadratic polynomial $P$, we know
$$P(x) = Q(x)(x-a)$$
$Q$ must be a first-degree polynomial, since anything higher-degree, multiplied by a first-degree polynomial, would produce a higher-than-second-degree polynomial.  So
$$P(x) = (x-b)(x-a)$$
Now, assuming we're working over an integral domain (which $\mathbb R$ and $\mathbb C$ both are),
$$P(x) = 0 \Rightarrow x-b=0 \text{ or } x-a=0$$
So $a$ and $b$ are the only zeros of $P$ (although it is possible that $a=b$).
A: Suppose not. Then there are at least 3 roots and so $$P(x)=(x-x_1)(x-x_2)(x-x_3)Q(x)$$ This is at least a cubic so this cannot happen.
A: $$0 = ax^2 + bx + c$$
We solve this equation by completing the square. It offers up to two distinct solutions. The name we give to the general solutions is the quadratic formula. That's all there is.
If we consider the case of real solutions, and you think there may be a sneaky third solution, remember that $f(x) = ax^2 +bx +c$ can be plotted as below (depending on the sign of $a$). How many times could a parabola cross a horizontal line?

A: A more general answer to this question lies in the following theorem:
Theorem If $P(x)$ is a polynomial of degree $n$, and $a$ is a value for which $P(a) = 0$, then $P(x) = (x - a)Q(x)$, where $Q(x)$ is a polynomial of degree $n - 1$.
This theorem is a simple consequence of polynomial long division. By long division, $P(x) = (x - a)Q(x) + R(x)$, for some polynomials $Q(x), R(x)$ with the degree of $R(x)$ less than the degree of $(x-a)$. But since $x - a$ is of degree 1, that means $R(x)$ is of degree $0$. I.e., $R(x) = R$, a constant.
But $P(a) = 0$, so $0 = (a - a)Q(a) + R$, and so $R = 0$ and we get just $P(x) = (x-a)Q(x)$. Since the degree of the product of two polynomials is the sum of their degrees, the degree of $P(x)$ is one greater than that of $Q(x)$, so the degree of $Q(x)$ must be $n-1$.

Now, if $P_n(x)$ is of degree $n > 0$ and $a_n$ is a root, then $$P_n(x) = (x - a_n)P_{n-1}(x)$$ for some $n-1$ degree polynomial $P_{n-1}(x)$. If $P_n(x)$ has another root $a \ne a_n$, then $a$ must also be a root of $P_{n-1}(x)$:
$$0 = P_n(a) = (a - a_n)P_{n-1}(a)$$
Since $a - a_n \ne 0$, we can divide it out to get $P_{n-1}(a) = 0$.
Conversely, if $a_{n-1}$ is a root of $P_{n-1}$, then $$P_n(a_{n-1}) = (a_{n-1} - a_n)P_{n-1}(a_{n-1}) = 0$$
So $a_{n-1}$ must also be a root of $P_n$ (which may be the same or different from $a_n$). We can also apply the theorem to $P_{n-1}$ and $a_{n-1}$:
$$P_{n-1}(x) = (x - a_{n-1})P_{n-2}(x)$$
for some degree $n-2$ polynomial $P_{n-2}(x)$. By combining, we see that $$P_n(x) = (x - a_n)(x - a_{n-1})P_{n-2}(x)$$
As long as we can keep finding roots for the reduced polynomials, we can keep this up. If we can find $k$ such roots,
$$P_n(x) = (x - a_n)(x - a_{n-1})(x - a_{n-2})...(x - a_{n+1-k})P_{n-k}(x)$$
Then $P_{n-k}(x)$ has to be a polynomial of degree $n-k$.
If we can find $n$ such roots, then 
$$P_n(x) = (x-a_n)(x-a_{n-1})...(x-a_1)P_0$$
where $P_0$ is a constant (a $0$-degree polynomial). $P_0 \ne 0$, since if it were we would have $P_n(x) = 0$ everywhere. But then the degree of $P_n$ would be $0$ (or less - some people define the degree of the $0$ to be $-\infty$), contrary to our original condition on $P_n(x)$. So in this case, $P_n(x)$ cannot have any other roots distinct from $a_1, a_2, ..., a_n$, since any other value would leave all factors in the expression non-zero.
So $P_n(x)$ can have at most $n$ roots.
The Fundamental Theorem of Algebra says that any non-constant polynomial over the complex numbers has a root. This theorem requires a substantial development of the properties of complex numbers to prove. But by it, we see that the process above does not terminate until you get to the constant. Thus a polynomial of degree $n$ will always have exactly $n$ roots $a_1, a_2, ..., a_n$. But remember that the $a_i$ values do not have to be distinct. The number of times a particular value occurs in this list is called the multiplicity of the root. So you only get $n$ if you count the roots by their multiplicity.
A: Hint $ $ Suppose $f(x)\,$ is a polynomial of $\color{#0a0}{{\rm degree}\,2}\,$ with coef's in a field (or domain) $F$ (e.g. $\,\Bbb Q,\Bbb R,\Bbb C)$ and suppose that $\,f\,$ has $\,2\,$ distinct roots $\,a\neq b.\,$ By the Bifactor Theorem below we deduce that $\,f(x) = c(x\!-\!a)(x\!-\!b)\,$ for $\,\color{#0a0}{0\neq c}\in F.\,$ 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). Thus a $\rm\color{#0a0}{quadratic}$ has at most $\,\color{#0a0}2\,$  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 $\rm\: \color{#C00}{a\!-\!b\ne 0},$
$$\begin{eqnarray}\rm\:f(b)= 0 &\ \Rightarrow\ &\rm f(x)\, =\, (x\!-\!b)\,g(x)\ \ for\ \ some\ \ g\in F[x]\\
\rm f(a) = (\color{#C00}{a\!-\!b})\,g(a) = 0 &\Rightarrow&\rm g(a)\, =\, 0\,\ \Rightarrow\ g(x) \,=\, (x\!-\!a)\,h(x)\ \ for\ \ some\ \ h\in F[x]\\
&\Rightarrow&\rm f(x)\, =\, (x\!-\!b)\,g(x) \,=\, (x\!-\!b)(x\!-\!a)\,h(x)\end{eqnarray}$$
Remark $ $ More generally, by inductively iterating the Factor Theorem (as we did above) we deduce that a nonzero polynomial $\,f\,$ over a field (or domain) has no more roots than its degree $\,n.\,$  Indeed if $\,f\,$ has  $\,\ge n\,$ distinct roots $\,r_i$ then inductively applying the  Factor Theorem shows that $\,f = c(x\!-\!r_1)\cdots (x\!-\!r_n),\,$ so $\ r\ne r_i\Rightarrow\, f(r)= c(r\!-\!r_1)\cdots (r\!-\!r_n) \ne 0\,$ by all factors are $\ne 0.\,$ Thus $\,f\,$ has at most $\,n\,$ roots.
The above root-bound property characterizes (integral) domains (commutative rings $\ne \{0\}$ which satisfy $\rm\,ab=0\,\Rightarrow\, a=0\,$ or $\rm\,b=0),\,$  viz. a ring $\rm\: D\:$ is a domain $\iff$ every nonzero polynomial $\rm\ f(x)\in D[x]\ $ has at most $\rm\ deg\ f\ $ roots in $\rm\:D.\:$ For a simple proof see  this answer, where I illustrate it constructively in $\rm\: \mathbb Z/m\: $ by showing that, given any $\rm\:f(x)\:$ with more roots than its degree, we can quickly compute a nontrivial factor of $\rm\:m\:$ via a quick $\rm\:gcd.\,$
The quadratic case of this result is at the heart of some integer factorization algorithms, which e.g. attempt to factor $\rm\:m\:$ by searching for a square-root of $1$ that is nontrivial $(\not\equiv \pm1)$ in $\rm\: \mathbb Z/m.$
Beware that there are very simple examples of failure in non-domains, e.g. if $\,ab=0, a,b\neq 0\,$ then $\,ax\,$ has at least $2$ roots $\,b,0,\,$ and $\,(x-a)(x-b)\,$ has at least $\,4\,$ roots $\,a,b,0,a+b\, $ if $\,a\neq  b.\,$ A simple concrete case is in $\,\Bbb Z_8 = $ integers $\!\bmod 8\!:\,$ $\rm{odd}^2= 1\,$ so $\,x^2-1\,$ has $\,4\,$ roots $\,\pm1,\pm 3.$
A: Suppose there are three distinct roots $x,y,z$. One has
$$\begin{cases}ax^2+bx+c=0\\ay^2+by+c=0\\az^2+bz+c=0\end{cases}\Rightarrow\begin{cases}a(x^2-y^2)+b(x-y)=0\\a(x^2-z^2)+b(x-z)=0\end{cases}\Rightarrow\begin{cases}a(x+y)+b=0\\a(x+z)+b=0\end{cases}$$ It follows $$a(z-y)=0\Rightarrow z=y$$ which is a contradiction.
A: Suppose it has three roots and $a\neq0$.
The hypotheses of Rolles Theorem are satisfied then there will exist two roots of the derivative and a root of the second derivative which is a constant $(=2a)$.
A: Let
$$r=\frac{-b+\sqrt{b^2-4ac}}{2a},\quad s=\frac{-b-\sqrt{b^2-4ac}}{2a}.$$
Simple calculation shows that
$$r+s=-\frac ba\quad\text{ and }\quad rs=\frac{b^2-(b^2-4ac)}{4a^2}=\frac ca.$$
Thus
$$a(x-r)(x-s)=a[x^2-(r+s)x+rs]=ax^2+bx+c.$$
If $t$ is any root of the quadratic equation $ax^2+bx+c=0,$ then
$$a(t-r)(t-s)=at^2+bt+c=0.$$
Since $a\ne0$ this means that
$$(t-r)(t-s)=0$$
whence
$$t-r=0\quad\text{ or }\quad t-s=0,$$
i.e.,
$$t=r\quad\text{ or }\quad t=s.$$
A: If you wrote the theorem out, it would look like this
THEOREM: Let $a,b,$ and $c$ be real numbers with $a \ne 0$. Then $ax^2+bx+c=0$ if and only if $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. 
That means
if $ax^2+bx+c=0$, then $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ 
and
if $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, then $ax^2+bx+c=0$.
A: Suppose $a \neq 0$. Then $ax^2+bx+c=0$ has the same set of solutions as $x^2+\frac{b}{a}x+\frac{c}{a}=0$ which means that we can reduce all equations to
$$x^2+px+q=0$$
without losing generality.
Suppose $q=0$, then it is $x^2+px=0$ when either $x=0$ or $x+p=0$, but $x+p=0$ has only one solution by definition $x=-p$
Next, take then $q \neq 0$, when
$$x(x+p)=-q$$
Take $y=x+\frac{p}{2}$ which makes
$$(y+\frac{p}{2})(y-\frac{p}{2})=y^2-\frac{p^2}{4}=-q$$
or
$$y^2=-q+\frac{p^2}{4}=b$$
So it comes down to a question of how many different roots square root may have.
Write two of them which we suspect:
$$y_1^2=b$$
$$y_2^2=b$$
$$y_1^2-y_2^2=(y_1-y_2)(y_1+y_2)=0$$
So it is either $y_1=y_2$ which gives the same solution or $y_1=-y_2$ which connects the solutions in such a way that there cannot be two different negative values of the same number. Since the connections of reduced solution $y_1,y_2$ are connected with $y_1=-y_2$ the starting set of solutions must have the same cardinality: no more than $2$.
Finally if $a=0$ there can be only one solution by definition.
