$$\ x^3+3px=2q $$

Question

$$\ x^3+3px=2q $$

anonymous · Answer

$$x=\sqrt[3]{(q+ \sqrt{q^3+p^3})}-\sqrt[3]{(-q+ \sqrt{q^3+p^3})}$$

anonymous · Answer

From where I'm getting this: 'by considering the existence of stationary points it is easy to show that a cubic with real roots has $$q^3+p^3<0$$

anonymous · Answer

I don't get this. Surely that would make the square root expression imaginary, hence the roots of x would be complex?

zzr0ck3r · Answer

I have no idea, and dont call me Shurley.

anonymous · Answer

Here, in your solution, value of q is squared [not the cubic value] 
Then the correct answer is:
$$x=\sqrt[3]{(q+\sqrt{q ^{2}+p ^{3}})}-\sqrt[3]{(-q+\sqrt{q ^{2}+p ^{3}})}$$
So, the equation has three real roots when :
$$q ^{2}+p ^{3}<0$$

anonymous · Answer

But WHY does it have 3 REAL roots given that inequality?

anonymous · Answer

So, You Want to Solve the Cubic?: This lecture note explains the theory: http://www.math.dartmouth.edu/archive/m43s09/public_html/cubic-lecture.pdf

anonymous · Answer

Is what he calls x_k= the equation that you typed above?

anonymous · Answer

page 11

anonymous · Answer

@henpen yess:) u can see the proof in page 12-13.