Re: Why is the negative root?

*To*: mathgroup at smc.vnet.net*Subject*: [mg69608] Re: Why is the negative root?*From*: p-valko at tamu.edu*Date*: Sun, 17 Sep 2006 06:57:50 -0400 (EDT)*References*: <200609130803.EAA18412@smc.vnet.net><eee0fi$6gl$1@smc.vnet.net> <eegap1$gp$1@smc.vnet.net>

Paul Abbott wrote: > In this case, the single root can be represented by this radical. But > modify your example slightly: > Reduce[{z^3 - z^2 - b z + 3 == 0, b > 0, z > 0}, z] // FullSimplify > How would you prefer the result to be expressed now? The answer is: b > (-1 - 647/(50867 + 5904*Sqrt[82])^(1/3) + (50867 + 5904*Sqrt[82])^(1/3))/12 && ((Sqrt[1 + 3*b] + (2 + 6*b)*Cos[(Pi/2 - ArcTan[(-79 + 9*b)/(3*Sqrt[3]*Sqrt[-231 + 54*b + b^2 + 4*b^3])])/3])/ (3*Sqrt[1 + 3*b]) || (Sqrt[1 + 3*b] - (1 + 3*b)*Cos[(Pi/2 - ArcTan[(-79 + 9*b)/(3*Sqrt[3]*Sqrt[-231 + 54*b + b^2 + 4*b^3])])/3] + Sqrt[3]*(1 + 3*b)*Sin[(Pi/2 - ArcTan[(-79 + 9*b)/(3*Sqrt[3]*Sqrt[-231 + 54*b + b^2 + 4*b^3])])/3])/ (3*Sqrt[1 + 3*b])) The answer is in every engineering handbook. They call it the "Cardano formula". Regards, Peter

**References**:**Why is the negative root?***From:*p-valko@tamu.edu