       Re: Different (real) solutions using Solve for same equation

• To: mathgroup at smc.vnet.net
• Subject: [mg101847] Re: [mg101807] Different (real) solutions using Solve for same equation
• From: Murray Eisenberg <murray at math.umass.edu>
• Date: Sun, 19 Jul 2009 07:15:10 -0400 (EDT)
• Organization: Mathematics & Statistics, Univ. of Mass./Amherst
• References: <200907180848.EAA06417@smc.vnet.net>

```You have a typo in your second Solve, below.  Once you fix that, you
should obtain the same solutions:

old = Solve[w/p == 25 e^2/(e + w)^2, w];
new = Solve[w^3 + 2 e w^2 + w e^2 == 25 p e^2, w];
old == new
True

kristoph wrote:
> Dear all,
>
> I came across the following observation which I find troublesome.
>
> I was trying to solve the equation w/p == 25 e^2 / (e + w)^2 using
> Solve[w/p == 25 e^2 / (e + w)^2 , w]. But the non-complex solution did
> not have the properties I wanted.
>
> I tested whether the solution was right and tried solving w^3 + 2 e
> w^2 w e^2 == 25 p e^2 (which is just rewriting the first equation)
> using Solve[w^3 + 2 e w^2 w e^2 == 25 p e^2, w].
>
> This time the properties were present and the two non-complex
> solutions using the first and second approach where different. I would
> like to know why? Each equation can be transformed into the other via
> simple operations, why are there different solutions to it?
>
>

--
Murray Eisenberg                     murray at math.umass.edu
Mathematics & Statistics Dept.
Lederle Graduate Research Tower      phone 413 549-1020 (H)
University of Massachusetts                413 545-2859 (W)
710 North Pleasant Street            fax   413 545-1801
Amherst, MA 01003-9305

```

• Prev by Date: Re: Re: Re: Releasing memory
• Next by Date: Re: Determine if a parameter is a function
• Previous by thread: Different (real) solutions using Solve for same equation ?
• Next by thread: Re: Different (real) solutions using Solve for same equation ?