Cubic Poynomial : In the equation $x^3 +3Hx +G=0$ if G and H are real and $G^2 +4H^3 >0$ then roots of the......... Question: 
In the equation $x^3 +3Hx +G=0$ if G and H are real and $G^2 +4H^3 >0$ then roots of the equation are
(a) all real and equal 
(b) all real and distinct 
(c)  one real and two imaginary 
(d) all real 
What I did : 
Let the cubic polynomial is $ax^3+bx^2+cx+d=0$ then let p,q,r are roots of equation then $ax^3 - a(p+q+r)x^2 + a(pq+pr+qr)x - a(pqr)$
How do I relate this with the given equation.. thanks.. 

The answer is (c): The equation has one real root and two nonreal complex conjugate roots. 
Proof: This Wiki article classifies the nature of the roots.
The discriminant of a cubic equation $ax^3+bx^2+cx+d=0\;$ is
$$\Delta = 18 a b c d  - 4b^3d + b^2 c^2 -4ac^3 - 27 a^2 d^2.$$


*

*If $\Delta > 0,$ then the equation has three distinct real roots.

*If $\Delta = 0,$ then the equation has a multiple root and all its roots are real.

*If $\Delta < 0,$ then the equation has one real root and two nonreal complex conjugate roots.


With your coefficients we get  (since $b=0$)
$$\Delta = -4ac^3 - 27a^2d^2 = -108H^3-27G^2 = -27(4H^3 + G^2) < 0.$$
And therefore, because $\Delta < 0,$  the equation has one real root and two nonreal complex conjugate roots.
Note: If the question is a kind of multiple choice, the third roots of unity (1, and two conjugate-complex) would also give a hint for case (c).
