I want to prove, for the numbers a and b, that the following equation has exactly three solutions if and only if 4a^3 + 27b^2 < 0:
x^3 + ax + b = 0,
x in R
Because, f(x) = x^3 + ax + b
=> f'(x) = 3x^2 + a
f'(x) = 0 => x = +sqrt(-a/3) or x = -sqrt(-a/3)
f''(x) = 6x
f''(x=+sqrt(-a/3)) = ...
It is proved that there are a set number of solutions for a particular function.