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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
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