[Math] 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………

polynomials

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..

Best Answer

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).

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