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Re: principle root? problem

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

Mathematica interprets radicals using principal roots for
Power[whatever, 1/integer]. This is documented in The Mathematica Book
(p. 754 in version 4).

For your example, Mathematica actually gives an empty solution set;
x->256 is the solution for right-hand-side set to +16.

In[2]:= Solve[x^(1/2)==16]
Out[2]= {{x -> 256}}

In[3]:= Solve[x^(1/2)==-16]
Out[3]= {}

To see the solution obtained by clearing radicals (which was later found
to be wrong, and thus discarded), inhibit verification.

In[4]:= Solve[x^(1/2)==-16, VerifySolutions->False]
Out[4]= {{x -> 256}}

This use of principal roots is in fact at the root of the discrepancy
between what you get for your first system and what you may have
expected. Again this shows up by inhibiting verification.

In[5]:= Solve[(x^2-5)^(2/3)==16]
Out[5]= {{x -> -Sqrt[69]}, {x -> Sqrt[69]}}

In[7]:= Solve[(x^2-5)^(2/3)==16, VerifySolutions->False]
Out[7]= {{x -> -I Sqrt[59]}, {x -> I Sqrt[59]}, {x -> -Sqrt[69]}, 
	{x -> Sqrt[69]}}

Note that this tactic of stopping the verifier is quite dependent on
exactly what internal transformations took place in the solving process.
It could well happen that different approach might not generate the
parasite solutions. So this is not a guaranteed method if your interest
is in finding all possible parasites.

Daniel Lichtblau
Wolfram Research

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