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Re: principle root? problem

> Now, is this a
> "principle root" problem and are the two possibly related?

Yes to both questions; though you may have meant x^(1/2)= 16, not
x^(1/2)= -16 , which Mathematica says has no solution.

To Mathematica,  z^a , when z is not 0, is the principle value:

        Exp[a(Log[Abs[z] + I Arg[z])]

where Log[Abs[z]] is the ordinary logarithm of the real number Abs[z] and
Arg[z] is the principle argument of z, the one lying in (-Pi, Pi] (including
Pi but not -Pi).

One would automatically think of changing the first equation to

    Solve[(x^2 - 5)^2 == 16^3, x]

        {{x -> (-I)*Sqrt[59]}, {x -> I*Sqrt[59]},
          {x -> -Sqrt[69]}, {x -> Sqrt[69]}}

The following way of making Mathematica look at all values, not just the
principle value, works for rational powers:

    Solve[{u^2 == 16, u^3 == x^2 - 5}, x]

        {{x -> (-I)*Sqrt[59]}, {x -> I*Sqrt[59]},
          {x -> -Sqrt[69]}, {x -> Sqrt[69]}}

     Solve[{u == -16, x == u^2}, {u, x}]

           {{x -> 256, u -> -16}}


Allan Hayes
Mathematica Training and Consulting
Leicester UK
hay at
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565

"RDownes" <rdownes at> wrote in message
news:a2r4jh$9p4$1 at
> 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
> Also, Mathematica gives the solution to x^(1/2)= -16 as 256. Now, is this
> "principle root" problem and are the two possibly related? Any insights
> be appreciated for this little enigma.
> Thanks,
> Rob

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