Need help with a proofs question.. Q : If $x_1$ and $x_2$ are two solutions to quadratic equation $ax^2+bx+c = 0$, then show that
$$x_1 + x_2 = -\frac{b}{a}\qquad\text{and}\qquad x_1x_2 = \frac{c}{a}.$$
From this question I gathered that the discriminant $b^2-4ac > 0$
I also tried to use the AGM and I got the inequality of $$\frac{c}{a} < \frac{b^2}{4a^2}.$$ But I don't know if I'm on the right track or not. What should I do? 
 A: Here is a funny way to show that the sum $S=x_1+x_2= - \frac b a$
We have 
\begin{align} 
1)&\quad ax_1^2+bx_1+c = 0 \\
2)&\quad ax_2^2+bx_2+c = 0
\end{align}
Subtracting 1 and 2
$$ax_1^2+bx_1-ax_2^2-bx_2 = 0$$
$$(x_1-x_2 )(aS+b )= 0$$
Since $ x_1 \ne x_2 $ here then
$$\fbox {$S=- \frac b a$ }$$
A: hint
If $$\Delta=b^2-4ac>0$$ then
$$x_1=\frac {-b-\sqrt {\Delta}}{2a} $$
and
$$x_2=\frac {-b+\sqrt {\Delta}}{2a} $$
the sum gives
$$ (x_1+x_2)=\frac {-b-\sqrt {\Delta}-b+\sqrt {\Delta}}{2a} $$
$$=\frac {-2b}{2a}=-b/a $$
and the product
$$x_1x_2=\frac {(-b-\sqrt {\Delta})(-b+\sqrt {\Delta})}{4a^2} $$
$$=\frac {(-b)^2-\Delta}{4a^2}=\frac {4ac}{4a^2}=c/a $$
A: when you try to prove any thing try to think in the simple way first and start with what you already know.
$$x_1 = \frac{- b + \sqrt{b^2 - 4ac}}{2 a}$$
$$x_2 = \frac{- b - \sqrt{b^2 - 4ac}}{2 a}$$
Thus
$$x_1 + x_2 = \frac{-2b}{2a} = -\frac{b}{a};$$
the square root part cancel out because the had different signs.
\begin{align}
x_1  x_2
&= \frac{(- b + \sqrt{b^2 - 4ac})}{2  a}  \frac{(- b - \sqrt{b^2 - 4ac})}{2  a} \\
&= \frac{b^2 - b^2 + 4  a  c}{4  a^2} \\
&= \frac{c}{a}.\end{align}
A: Let's call your original polynomial $P(x)$:
$P(x) = ax^2 + bx + c$
Now consider the polynomial
$Q(x) = a(x-x_1)(x-x_2) = a x^2 - a(x_1+x_2)x + a x_1 x_2$.
Hmm, both of them have the same two roots $x_1$ and $x_2$, and the same leading term $ax^2$.  Are they actually the same polynomial?  Let's look at the difference between them.
$D(x) = P(x) - Q(x) = (b + a(x_1+x_2))x + c - a x_1 x_2$
Since $x_1$ is a root of both $P$ and $Q$, we know $P(x_1)=Q(x_1)=0$, so
$D(x_1) = P(x_1)-Q(x_1) = 0-0 = 0$.
Similarly, $P(x_2)=Q(x_2)=0$, and
$D(x_2) = P(x_2)-Q(x_2) = 0-0 = 0$.
So $D(x)$ is a linear polynomial with value $0$ at two different points.  Its slope must be $(0-0)/(x_2-x_1)=0$, and its $y$-intercept must also be $0$.  So in fact $D$ is the zero polynomial, and $P=Q$.
(Note: It's more generally true that if two polynomials over the complex numbers have the exact same complex roots, including orders of any repeated roots, then one polynomial is a constant times the other polynomial.  But I didn't want to assume you'd come across this theorem.)
We determined the coefficients of $D$ in terms of $a$, $b$, $c$, $x_1$, and $x_2$ above, but now we know these coefficients are both zero:
$b + a(x_1 + x_2) = 0 \;\Rightarrow\; x_1 + x_2 = -\frac{b}{a}$
$c - a x_1 x_2 = 0 \;\Rightarrow\; x_1 x_2 = \frac{c}{a}$
