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MathGroup Archive 2002

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Re: principle root? problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg32563] Re: principle root? problem
  • From: Erich Neuwirth <erich.neuwirth at univie.ac.at>
  • Date: Sun, 27 Jan 2002 03:28:51 -0500 (EST)
  • References: <a2r4jh$9p4$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

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

--
Erich Neuwirth, Computer Supported Didactics Working Group
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