Solving Third-Degree EquationsDate: 09/24/2001 at 11:03:28 From: De Kort Subject: Re: Solve third-degree equations I'm trying to solve x^3 + 2x^2 - 3x - 4 = 0 p = -13/3 q = -38/27 To calculate t I need to take the square root of 4p^3 + 27q^2, which is -272... Of course I can consider t and u to be complex numbers, but in the end I'm looking for the r e a l solutions: and there are three of them! How can I solve this? Thanks, Jan Date: 09/25/2001 at 08:44:15 From: Doctor Jerry Subject: Re: Solve third-degree equations Hi Jan, To refresh my memory of this I looked at Morris Kline's _Mathematical Thought from Ancient to Modern Times_, page 266. Here's a sketch of Cardan's method - as completed by Euler. Start with the polynomial equation x^3 + a*x^2 + b*x + c = 0. Change variables by letting x=y - a/3. The polynomial equation becomes (1) y^3 + p*y + q = 0, where p = b - a^2/3 and q = c - a*b/3 + 2a^3/27. Equation (1) can be solved by letting y = u + v, thus introducing, for a moment, the two new unknowns u and v. Upon substitution into the equation and rearranging the result, we find the equation (2) u^3 + v^3 + (p + 3*u*v)(u+v) + q = 0. One condition on the two unknowns u and v is the requirement that (3) p + 3*u*v = 0, which simplifies (2). Equation (2) is now (4) u^3 + v^3 + q = 0. Solving (3) for v in terms of u, substituting the result into (4), and multiplying by u^3 gives, (5) (u^3)^2 + q(u^3) - p^3/27 = 0. Solving (5) by the quadratic formula, we find the roots A = -q/2 + (q^2/4 + p^3/27)^{1/2} and B = -q/2 - (q^2/4 + p^3/27)^{1/2}, where (q^2/4 + p^3/27)^{1/2} denotes either of the two square roots of q^2/4 + p^3/27. Equation (5) is satisfied if we take u^3 = A and v^3 = B. Letting A^{1/3} denote a fixed (but arbitrary) cube root of A, then the three possible solutions of (5) are (5) u_1 = A^{1/3}, u_2 = w*A^{1/3}, u_3 = w^2*A^{1/3}, where w = (-1 + i*(1/2)3^{1/2}, one of the cube roots of 1. If we choose v_1 = B^{1/3} as that cube root of B for which u_1*v_1 = -p/3, then corresponding to u_1, u_2, and u_3 will be v_1=B^{1/3}, v_2 = w^2*B^{1/3}, and v_3 = w*B^{1/3}. Equation (1) will then have the three roots y_1 = u_1 + v_1, y_2 = u_2 + v_2, and y_3 = u_3 + v_3. Finally, the roots to the equation x^3 + a*x^2 + b*x + c = 0 can be found easily. The "irreducible case" is the case in which q^2/4 + p^3/27 < 0. It can be shown that in the irreducible case all roots of (3) are real, although this is not obviously true from (5). There is a method due to Vieta, which simplifies the calculations in the irreducible case. I'll use the cubic x^3 + 2x^2 - 3x -4 = 0 you gave to illustrate. We first replace x by y - 2/3. This gives y^3 - (13/3)y - 38/27 = 0. Consider the trigonometric identity cos(3t) = 4*cos^3(t) - 3 cos(t) or, rearranging cos^3(t) - (3/4) cos(t) -(1/4)cos(3t) = 0. Letting y = n*z, the equation y^3 - (13/3)y - 38/27 = 0 becomes z^3 - (13/(3n^2)) z - 38/(27n^3) = 0. We try to make this like the rearranged trig identity. Let - (13/(3n^2)) = -3/4 and -38/(27n^3) = -(1/4)cos(3t). So, n = 2*sqrt(13)/3 and cos(3t) = 38/(26*sqrt(13)). We find t1 = 0.38447536423. Calculating t2 = t1+2pi/3 and t3 = t1+4pi/3 we find roots y1 = n*z1 = n*cos(t1) = 2.22821947948 y2 = n*z2 = n*cos(t2) = -1.8948864614 y3 = n*z3 = n*cos(t3) = -0.33333333333. Of course, we could have noticed (using the rational roots theorem) that -1 was a root of the original equation. Using synthetic division to find the quadratic factor, we find the other roots to be -1/2 +- sqrt(17)/2 . - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/ |
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