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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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