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. Help!
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 are: x = 4 x = -2+sqrt(3) x = -2-sqrt(3) -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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