^{3}+ax

^{2}+bx+c=0 to a simplier form, x

^{3}+Bx+C=0, by subsitiuting x=y-y

_{0}and choosing y

_{0}so that the coefficient of the squared term is zero. There are three roots for this equation and one can use complex numbers to see how they are related. Note that the sum of the three roots is zero.

Knowing the form of the solutions one can choose the components of the complex number x=r'+iq' and find the equation for which these numbers are the roots of. The equation below is the one that Cardan uses in Ars Magna.

The trick used to solve the simplified cubic equation involves substituting x=u-v and finding an expression relating u and v then using this to eliminate v and obtain a quadratic equation that is a function of u

^{3}. One can then use the quadratic formula to find u and the corresponding real root. The other roots can be found by multiplying u by the cube roots of 1 and solving for the root using the same method. The second solution for the quadratic equation involving u yields the same set of roots for the simplified cubic equation.

Supplemental (Mar 2): The above method for finding the roots of a cubic equation appears to work even if all the roots are real. For some choices of B the radical is imaginary and the associated roots end up as reals. For a double root (two roots the same) the radical is zero and for a triple root (all three roots the same) both B and C of the reduced equation are zero.

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