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Extraneous Solutions

Date: 08/08/2003 at 18:13:01
From: Jeffrey D. Grubs
Subject: Extraneous Solutions, and interpretation of their meaning

What should an engineer interpret from an equation, or the solutions 
of that equation, should its solutions be extraneous?

Example: Solve for x

(2/(x+1)) - (1/x) = (2/(x^2 + x) )

Solving for 'x' we find an answer of -2, yet checking the solution in 
the original equation results in a division by zero. 

As an engineer, I need to know what should be interpreted from the 
fact that the solution does not work with the equation. What does 
this mean about the equation being used or the extraneous solution? 
What would the interpretation be if we employed a function in which a 
single or range of values resulted in extraneous solutions? If we 
were employing a function or an equation in a mechanical or 
electrical application, what meaning should we give to the equation 
or the extraneous solutions derived from it?

The example in question was derived from a textbook example on 
extraneous solutions. I have been attempting to find some meaning 
that will allow me to derive the best possible handling of an 
engineering problem should an extranous solution result from formulas 
important to the field at hand. I am an engineer with a bachelors in 
electrical engineering and this issue has me concerned about being 
able to solve time dependent problems. 

Thank you for your help.


Date: 08/08/2003 at 23:50:51
From: Doctor Warren
Subject: Re: Extraneous Solutions, and interpretation of their meaning

Hi Jeffrey,

I'm a little confused by your question.  The equation

   2     1      2
  --- - --- = -----
  x+1    x    x^2+x

has only one solution, x = 3.  x = -2 is not a solution. Furthermore,
plugging in x = -2 does not result in a divide-by-zero, but instead
just results in the false statement

      2     1      2                 1
   - --- + --- != ---    or   - 2 + --- != 1
      1     2      2                 2

where '!=' denotes 'does not equal.'

The term 'extraneous solution' is usually applied to situations when
you square both sides of an equation; doing so often introduces new
roots, in addition to the roots of the original (un-squared) equation.
Here's an example from our archives:

   Why Multiple Roots?
   http://mathforum.org/library/drmath/view/52624.html 

Sometimes in physical situations, the term 'extraneous solution' will
be used to describe a _mathematically valid solution_ that is not
_physically_ valid. For example, an equation may have two solutions,
one of which represents a projectile moving above ground, and another
below ground; the below-ground solution would be taken to be 
unphysical.

A classic example of this sort of situation was Paul A. M. Dirac's
discovery of antimatter. He was working with equations whose solutions 
were also solutions of the relativistic mass-energy relationship

   E^2  = p^2 c^2  + m^2 c^4

where E is energy, p is momentum, m is rest-mass, and c is the speed
of light in a vacuum. If you take the square root of both sides,
you'll get two solutions for any given pair (p, m). For a very long
time, people threw out the negative-energy solution as being
unphysical - but Dirac challenged that notion, realizing that
"antimatter" particles were just what were needed to complete his
theory, which was an early attempt to unify quantum mechanics and
relativity theory. Positrons, the antimatter analogue of electrons,
were detected a short time later; antimatter particles now comprise
(exactly) half of the zoo of particles in the so-called Standard Model
of physics.

If you have any more questions about your function, or about
extraneous solutions in general, don't hesitate to ask.

- Doctor Warren, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Physics
High School Basic Algebra
High School Physics/Chemistry
Middle School Algebra

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