Solving Cubic and Quartic PolynomialsDate: 04/30/98 at 17:53:13 From: E. Aevan vonBerwick, IV Subject: Cubic, Quartic Formulas Hi, I know you have received numerous questions about the Cubic and Quartic Formulas, and I know they are very long, so I will not ask you to give them to me. But, I was wondering if you could tell me where I can find these formulas in their entirety, whether they be web pages or other math publications. Thank you, -Aevan Date: 04/30/98 at 19:02:58 From: Doctor Barney Subject: Re: Cubic, Quartic Formulas You're right, it would be difficult to write them in a text message. There is a very common handbook called the CRC Standard Mathematical Tables. It should be available in the reference section of any college library. Look in the index under "cubic." -Doctor Barney, The Math Forum Check out our web sites! http://mathforum.org/dr.math/ Date: 04/30/98 at 19:34:15 From: Doctor Anthony Subject: Re: Cubic, Quartic Formulas What you are wanting is Tartaglia's Solution (also known as Cardan's Solution) to the general cubic. It was invented by Tartaglia but was first published in 1545 by Cardan, who had obtained it from Tartaglia after giving a solemn promise not to reveal it. The general cubic ax^3 + 3bx^2 + 3cx + d = 0 is changed to a simpler cubic without a term in x^2 by the substitution: y = ax+b (or x = (y-b)/a) to get: y^3 + 3Hy + G = 0 where: H = ac-b^2 and G = a^2*d - 3abc + 2b^3 We now make use of the identity shown below, where w is a complex cube root of unity: x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x+wy+w^2*z)(x+w^2*y +wz) This is easily proved if you know how to factorize determinants, but if not, you can prove it the long way by multiplying out the factors, and remembering that w^3 = 1. The cubic is now expressed as: y^3 - 3pqy + p^3 + q^3 = 0 with: H = -pq and G = p^3 + q^3 So y^3 - 3pqy + p^3+q^3 = (y+p+q)(y+wp+w^2*q)(y+w^2p+wq) = 0 and the three roots will be: (1) y = -p - q (2) y = -wp - w^2q (3) y = -w^2p - wq So all that is required is that we have the values of p and q from the two equations: H = -pq and G = p^3 + q^3 Here, think in terms of p^3 and q^3 rather than p and q, and consider instead the two equations: p^3 + q^3 = G and p^3*q^3 = -H^3 Now we can see that p^3 and q^3 are the roots of the quadratic: t^2 - Gt - H^3 = 0 and their cube roots must be chosen so that pq = -H. The roots of this quadratic are: (1/2)[G +-sqrt(G^2 + 4H^3)] If G^2 + 4H^3 > 0 p and q are real and distinct If G^2 + 4H^3 = 0 p and q are real and equal. The corresponding roots of the cubic are -2p, p, p. If G^2 + 4H^3 < 0 p and q are complex. The cubic then, ironically, has 3 real roots. In this situation it is better to use a trigonometrical method. Trig Solution of Cubic Equation ------------------------------- Having expressed the equation in the form y^3 + 3Hy + G = 0 and finding that G^2 + 4H^3 < 0, we know that there are three real roots. For this method we use the trig identity: cos(3*theta) = 4*cos^3(theta) - 3*cos(theta) The method is best illustrated by an example: x^3 - 6x - 4 = 0 We note that: 3H = -6, H = -2, and G = -4 and: G^2 + 4H^3 = 16 - 32 = -16 so the condition for 3 real roots applies. Substitute in x = k*cos(theta) and choose k so that the left hand side is a multiple of 4*cos^3(theta) - 3*cos(theta): k^3.cos^3(theta) - 6k.cos(theta) = 4 This means that: k^3 4 ----- = --- 6k = 3 So k^2/6 = 4/3, which implies k^2 = 8. Thus, k = 2*sqrt(2). Solving for theta: 16*sqrt(2)cos^3(theta) - 12*sqrt(2)cos(theta) = 4 4*cos^3(theta) - 3*cos(theta) = 1/sqrt(2) cos(3*theta) = 1/sqrt(2) 3*theta = pi/4, 7pi/4, 9pi/4 theta = pi/12, 7pi/12, 3pi/4 Thus: x = 2*sqrt(2)cos(pi/12) or x = 2*sqrt(2)cos(7*pi/12) or x = 2*sqrt(2)cos(3*pi/4) = 2*sqrt(2)(-1/sqrt(2)) = -2 Having found that x = -2 is a root we could, if we wished, divide out by x+2 and then solve the resulting quadratic, to get the two other roots already found above. In fact: x^3 - 6x - 4 = (x+2)(x^2 - 2x - 2) = 0 So x = -2, and x = 1 +-sqrt(3) ------------------------------------------------------------------- QUARTIC I will outline the method for solving the general quartic: z^4 + 4pz^3 + 6qz^2 + 4rz + s = 0 As for the cubic, the first step is to reduce the equation to one with no term in z^3. So we get: w^4 + aw^2 + bw + c = 0 Use the identity: (w^2 + t)^2 = u(w + v)^2 (Equation 1) and show that t is a root of the CUBIC equation (known as the reducing cubic): 8t^3 - 4at^2 - 8ct + 4ac - b^2 = 0 If one of the roots is t and the corresponding values are then u and v from the identity (1) above, then: w^4 + aw^2 + bw + c = [w^2-u^(1/2)*w+t-vu^(1/2)][w^2+u^(1/2)*w+t+vu^(1/2)] putting each quadratic factor separately equal to 0 gives the values of w. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/