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Re: principle root? problem


if you solve

(x^2-5)^2=16^3
mathematica will give you 4 solutions, and this makes sense.

a^b with noninteger b is restricted to the positive reals for a



RDownes wrote:
> 
> The other day, I was explaining something to a student regarding solving a
> simple algebra problem.
> 
> (x^2-5)^(2/3)=16
> 
> The solution to which is easily found. All four that is!  However my version of
> Mathematica only gave the two real.  Is there a simple explanation for this?
> 
> Also, Mathematica gives the solution to x^(1/2)= -16 as 256. Now, is this a
> "principle root" problem and are the two possibly related? Any insights would
> be appreciated for this little enigma.
> 
> Thanks,
> 
> Rob

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