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MathGroup Archive 2002

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Re: principle root? problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg32535] Re: [mg32532] principle root? problem
  • From: Andrzej Kozlowski <andrzej at platon.c.u-tokyo.ac.jp>
  • Date: Sat, 26 Jan 2002 04:07:52 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

Yes, these phenomena are related. The solutions Mathematica gives are 
correct according to the usual conventions. You can see what happens as 
follows. First replace your equation by the equation:
(x^2 - 5)^2 == 16^3. You will get, as you wrote, four roots

In[1]:=
sols = Solve[(x^2 - 5)^2 == 16^3, x]

Out[1]=
{{x -> (-I)*Sqrt[59]}, {x -> I*Sqrt[59]}, {x -> -Sqrt[69]},
   {x -> Sqrt[69]}}

Now substitute the solutions into (x^2-5)^(2/3)-16

In[2]:=
Simplify[(x^2 - 5)^(2/3) - 16 /. sols]

Out[2]=
{16*(-1 + (-1)^(2/3)), 16*(-1 + (-1)^(2/3)), 0, 0}

In[3]:=
N[%]

Out[3]=
{-23.999999999999993 + 13.856406460551018*I,
   -23.999999999999993 + 13.856406460551018*I, 0., 0.}

As you see two of your "roots" do not satisfy the original equation and 
the two equations are not equivalent over the complex numbers. The 
reason is, of course, that (-1)^(2/3) is not 1. The reasons for this 
have been explained in this group a lot of times, so I suggest that if 
you are still not satisfied with this explanation you search the list 
archive.

Andrzej Kozlowski
Toyama International University
JAPAN
http://platon.c.u-tokyo.ac.jp/andrzej/



On Friday, January 25, 2002, at 04:58  PM, 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
>
>
>
>
>



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