Question

If the equation x^3 + px + q = 0 has real roots then prove that 27q^2 + 4p^3 < 0.

Answer 1

@inkyvoyd

Answer 2

lool I'm sleepy?

Answer 3

3 real roots?

Answer 4

YEp.

Answer 5

Start by reading wikipedia xD

Answer 6

there are well known formulas for the solution of a "reduced" cubic equation (which is what you have).
see here for example: http://www.trans4mind.com/personal_development/mathematics/polynomials/cubicAlgebra.htm

Answer 7

No. It must have a more direct solution.

Answer 8

you study dirivative?

Answer 9

Yep.

Answer 10

i think f(x)=x^3+px+q
find solution f'(x)=0 is x1 anh x2
f(x1)*f(x2)<0

Answer 11

Are you getting the req. expression?
Cause Im not.

Answer 12

????

Answer 13

A cubic
\[a x^3+b x^2+c x+d=0
\]
has three real roots if the discriminant
\[
\Delta =-27 a^2 d^2+18 a b c
d-4 a c^3-4 b^3 d+b^2 c^2 > 0
\]
in our case
a = 1; b = 0; c = p; d = q;
then
\[\Delta=-4 p^3-27 q^2
\]
we are done

Answer 14

@eliassaab , nice job, but where'd you get the discriminant and the has three real roots from?

Answer 15

http://en.wikipedia.org/wiki/Cubic_function

Answer 16

Damn that formula. Nice work.