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? 

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}
