f(x)=x3+px+qfind the condition on p and q so that f(x) has 3 distinct real roots..please reply as early as possible

Suppose that f(x) has three distinct roots. Then are are x1 and x2 (by Rolle's theorem) sitting between these roots such that f'(x1) = f'(x2) = 0:

Since f'(x) is a quadratic with roots x1 and x2; it follows that p < 0: Setting f0(x) = 0

we get  x1 := -(p/3)1/2 and x2 = -(p/3)1/2
From the second derivative we see that f has a local maximum and a local minimum at x1 and x2; respectively. Therefore f(x1) > 0 and f(x2) < 0; that is,

f(x1)f(x2) < 0        which gives 27q2 + 4p3 < 0

Conversely, suppose that 27q2 + 4p3 < 0: Then obviously p < 0: Therefore f'(x) = 0 has two roots x1 and x2 at which f has a local max and a local min. That there is a root between x1 and x2 follows from the fact that f(x1)f(x2) < 0

(so apply IVT). Since f(x) --> - infinity as x --> - infinity and  f(x) -->  infinity as x --> infinity; it follows that f has one root in
( - infinity, x1) and another in (x2, infinity):

Hence f has three roots.

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regards

Ramesh