Extraneous SolutionsDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/