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MathGroup Archive 1997

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Re: Algebra Problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg7649] Re: [mg7624] Algebra Problem
  • From: Richard Finley <trfin at fiona.umsmed.edu>
  • Date: Tue, 24 Jun 1997 03:36:13 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

Robert,

You hit the nail on the head for the cause of the problem...(-1)^(1/3) has
three solutions only one of which is real and solves your equation, so Mma
cannot assume which one you want.  If you first load: 

<<Miscellaneous`RealOnly`

and then look at your equation

-9 + 9 (-1)^(1/3) + 18 (-1)^(2/3)

Mma will now simplify correctly to 0, which shows it is the solution you
want. In addition, if you insert (-1)^(1/3) it will simplify it to -1 since
it can now ignore the complex solutions.  The problem is, if you do your
original Solve[ ... ] you still only get the first solution.  A better way
to proceed is to recognize from the beginning that your equation is simply a
quadratic equation (in the variable x^(1/3) ) so there must be 2 solutions,
substitute /. x -> u^3, Solve the resulting equation for u (which gives the
solutions { u -> -3 }, { u -> 3/2 }, and then simply cube the results to get
your x = -27,  27/8.

Hope that helps...
RF
 
At 04:15 PM 6/20/97 -0400, you wrote:
>
>I am using Mathematica to study Algebra. When I solve the following
>equation from my textbook with Mathematica, it gives me this result:
>
>In[1]:=
>Solve[2x^(2/3) + 3x^(1/3) - 9 == 0]
>
>Out[1]=
>       27
>{{x -> --}}
>       8
>
>However, my textbook claims that -27 is also a solution. If I tell
>Mathematica to turn off its solution verification, it finds the other
>textbook solution:
>
>In[213]:=
>Solve[2x^(2/3) + 3x^(1/3) - 9 == 0, VerifySolutions -> False]
>
>Out[213]=
>                   27
>{{x -> -27}, {x -> --}}
>                   8
>
>When I ask Mathematica to substitute -27 for x in the equation,
>Mathematica only goes so far in simplifying the equation, but not far
>enough to determine if the left side is the same as the right side:
>
>In[2]:=
>2x^(2/3) + 3x^(1/3) - 9 == 0 /. x -> -27
>
>Out[2]=
>           1/3          2/3
>-9 + 9 (-1)    + 18 (-1)    == 0
>
>Now, I can see that the cube root of -1 is -1, and that the taking the
>square of -1 yields 1, and then taking the cube root of that also yields
>1. So the equation should simplify to:
>
>-9 - 9 + 18 == 0
>
>and then
>
>0 == 0
>
>which would indicate that -27 is indeed a solution. But I can't figure out
>how to get Mathematica to return similar results. Trying to force
>Mathematica to give a numerical answer, yields an imaginary, non-zero
>result:
>
>In[3]:=
>N[2x^(2/3) + 3x^(1/3) - 9 /. x -> -27]
>
>Out[3]=
>-13.5 + 23.3827 I
>
>which may explain why Mathematica rejects this solution. So, is
>Mathematica rejecting the solution -27 appropriately or not?
>
>Thanks in advance.
>
>======================================================================
>      Robert McNally <mailto:ironwolfNO at SPAMdangerousgames.com>
>    Visit <http://personalweb.lightside.com/pfiles/mcnally1.html>
>----------------------------------------------------------------------
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>



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