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 Visit our SunSITE at http://sunsite.univie.ac.at Phone: +43-1-4277-38624 Fax: +43-1-4277-9386