Fundamental Mathematics from an Advanced Standpoint

Math 4303

Helmut Knaust, Department of Mathematical Sciences, UTEP, El Paso TX 79968
hknaust@utep.edu

Last Edits: 3/1/2011

Cubic Equations and Complex Numbers

Below are Mathematica' s three solutions to the cubic equation (x^3)-2(x^2)-5x+7=0.
Mathematica seems to use Cardano' s Formula (wow - no progress since then?). 
Note the complex i under the cube roots, stemming from square roots of negative numbers.

Solve[x^3-2x^2-5x+7==0][[1,1,2]]
Solve[x^3-2x^2-5x+7==0][[2,1,2]]
Solve[x^3-2x^2-5x+7==0][[3,1,2]]

A plot of the poynomial reveals that the equation has three real solutions:

Plot[x^3-2x^2-5x+7,{x,-4,4}]

And in fact, here are numerical approximations:

NSolve[x^3-2x^2-5x+7==0][[1,1,2]]
NSolve[x^3-2x^2-5x+7==0][[2,1,2]]
NSolve[x^3-2x^2-5x+7==0][[3,1,2]]

Here is Cardano' s Formula in general.
 - It suffices to find one solution; the other two can the be found by polynomial division and the quadratic formula.
 - He showed first that it suffices to look at polynomials without quadratic term.

Solve[x^3+3b x+c==0,x][[1,1,2]]