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

Out=
{{x -> (-I)*Sqrt}, {x -> I*Sqrt}, {x -> -Sqrt},
{x -> Sqrt}}

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

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

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

In:=
N[%]

Out=
{-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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