Using Cardan's Formula to find Real RootsDate: 04/13/2000 at 07:21:40 From: Shadi R Salman Subject: Cardan's formula, complex numbers, and cubic equations How can I show that, although the cubic equation x^3 - 6x = 4 has three real solutions, Cardan's formula can find them by subtracting appropriate cube roots of complex numbers? I tried to solve by simply substituting in Cardan's formula (below): (((p/3)^3 + (q/2)^2)^(1/3) + q/2)^(1/3) - (((p/3)^3 + (q/2)^2)^(1/3) - q/2)^(1/3) where p = -6, q = 4 and obtained only one answer, which was complex. Hence, it is a wrong answer. The answer I got had sqrt[3]/2 - 1/2 + i(sqrt([3]/2 - 3/2) as the only root (and it's not even real). It doesn't even work in the original cubic. By using a trigonometric formula, I know that the three correct real roots are: x1 = (3)^0.5 + 1 = 2.732 x2 = -2 x3 = 1 - (3)^0.5 = 0.732 Please help. Date: 04/13/2000 at 08:11:10 From: Doctor Rob Subject: Re: Cardan's formula, complex numbers, and cubic equations Thanks for writing to Ask Dr. Math, Shadi. The problem is that q = -4, not +4. First you have to move all of the terms to one side of the equation: x^3 - 6*x - 4 = 0. Then you can read off p = -4 and q = -4. See if that helps. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ Date: 04/14/2000 at 04:57:02 From: Shadi R Salman Subject: Re: Cardan's formula, complex numbers, and cubic equations Using p = -6 and q = -4 did NOT help, for it also gives a complex answer. Even then, what puzzles me is, how am I supposed to get 3 answers from a single formula? Date: 04/14/2000 at 09:19:18 From: Doctor Rob Subject: Re: Cardan's formula, complex numbers, and cubic equations Yes, you are correct. The formula above does apply with q = 4. One problem is that the formula is not correct as written. The expression ((p/3)^3 + (q/2)^2)^(1/3) should be instead ((p/3)^3 + (q/2)^2)^(1/2) that is, a square root, not a cube root. If you fix that and substitute p = -6 and q = 4, you'll see that this expression becomes 2*i. Then the root given by the formula is x = (2*i+2)^(1/3) - (2*i-2)^(1/3) Now you have to consider which of the three cube roots each of these complex numbers to use. One choice is (2*i+2)^(1/3) = -1 + i and (2*i-2)^(1/3) = 1 + i They lead to the root x2 = -2. Other choices will lead to the other roots. The other cube roots are found by multiplying those above by (-1+sqrt[-3])/2 and (-1-sqrt[3])/2 the two complex cube roots of 1. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ Date: 04/14/2000 at 05:11:58 From: Shadi R Salman Subject: Re: Cardan's formula, complex numbers, and cubic equations Sorry, the question posed does have the wrong formula. However, the corrected question is as follows: In the form of x^3 + px = q Cardan's formula for the solution of the cubic equation is (((p/3)^3 + (q/2)^2)^(1/2) + q/2)^(1/3) - (((p/3)^3 + (q/2)^2)^(1/2) - q/2)^(1/3) How can I show that, although the cubic equation x^3 - 6x = 4 (i.e. p = -6, q = 4) has three real solutions, Cardan's formula can find them by subtracting appropriate cube roots of complex numbers? The real question is: How can I get three answers from that one formula? When I tried to solve by substituting in Cardan's formula (corrected), I still obtained only one answer, which was complex. Hence, it is a wrong answer. The answer I got had -(sqrt[3])/2 - 1/2 + i(sqrt[3]/2 - sqrt[3]/2 as the only root (again, it's not even real and it doesn't work in the original cubic equation either). Please help. Date: 04/14/2000 at 05:46:40 From: Doctor Anthony Subject: Re: Cardan's formula, complex numbers, and cubic equations I have given a general note on Cardan and on the Trig solution (when there are 3 real roots.) Cardan is VERY unwieldy when you have 3 real roots because the quadratic in t has complex roots. I hope you will be able to sort out where the difficulty is. Note too that what I have called p and q are NOT the same as in your equation. Look at: http://mathforum.org/dr.math/problems/vonberwick4.30.98.html - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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