ab > 0 and bc < 0
case 1 ...
a and b both > 0 ... then c < 0
then ac < 0, -4ac > 0
b^2 - 4ac > 0 ... two distinct real roots
case 2 ...
a and b both < 0 ... then c > 0
ac < 0 , -4ac > 0
b^2 - 4ac > 0 ... two distinct real roots
If ab>0 and bc<0, then ax^2+bx+c=0 has two real solutions. (Prove this)
I started, but I don't know.
Let a, b, c be real.
Assume ab>0 and bc<0.
So bc<ab
So c>a.
So....
I don't know from here. It clearly involves showing that the discriminant, b^2-4ac is positive I would think. How I can go about doing so, not sure.
Thanks for the help.