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Roots of a Cubic Equation

Date: 05/14/97 at 18:51:13
From: victoria givins
Subject: Complex number's cubic roots

I need to know how to solve the cubic x^3 = 15x+4 using Cardona's 
formula. I know that x = t^1/3+u^1/3 is a root of the equation if 
t+u = -b and t^1/3u^1/3 = -(a/3). I also know that t = 2+(-121)^1/2 
and u = 2-(-121)^1/2.  


Date: 05/15/97 at 09:57:53
From: Doctor Anthony
Subject: Re: Complex number's cubic roots

Tartaglia's solution has the following form.  First express the 
equation as:

    x^3 + 3Hy + G = 0      

In your example, x^3 + 3(-5)x +(-4) = 0 and we have H = -5, G = -4.

To give the three roots, we then use the identity:

x^3 - 3pqx + p^3+q^3 = (x+p+q)(x+wp+w^2*q)(x+w^2*p+wq) = 0

The complex cube roots of unity are w and w^2.

i.e.  x = -p-q
      x = -wp-w^2*q
      x = -w^2*p-wq

If p, q are chosen so that p^3 + q^3 = G and pq = -H, then p^3, q^3 
will be the roots of the quadratic: 

        t^2 - Gt - H^3 = 0

Their cube roots must then be chosen so that  pq = -H.

The roots of the quadratic are (1/2)[G +- sqrt(G^2+4H^3)].

If G^2+4H^3 > 0, p and q are real, so the roots of the cubic are one 
real and two complex conjugates.

If G^2+4H^3 = 0, p and q are real and equal.  The cubic equation has 
three roots: -2p, p, p.

If G^2+4H^3 < 0, p and q are complex and this will mean that the cubic 
has 3 real roots.  The equation is best solved by trig substitution.

For our example, G^2+4H^3 = (-4)^2 + (-5)^3  = 16 - 125 = -109

So this means that the cubic has three real roots, and indeed they 

    x = 4
    x = -2+sqrt(3)
    x = -2-sqrt(3)

-Doctor Anthony,  The Math Forum
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Associated Topics:
High School Polynomials

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