Jump to content

Talk:Extraneous and missing solutions

Page contents not supported in other languages.
From Wikipedia, the free encyclopedia
This is an old revision of this page, as edited by Tparameter (talk | contribs) at 07:15, 19 January 2008 (First Example). The present address (URL) is a permanent link to this revision, which may differ significantly from the current revision.

First Example

Wow. The first example in this article is absolutely terrible. It's obvious at a glance that solutions where x=2 or x=-2 are not valid. After deriving x=-2, we have the statement, "We arrive at what appears to be a solution rather easily. However, something very strange occurs when we substitute the solution found back into the original equation..." Actually, nothing strange occurs at all. The solution is not part of the domain of potential solutions - which should be obvious before starting. {sigh}. Of course, the second example is even worse. I would personally improve this article if I had any idea what the point of it is. Anyone else? Tparameter (talk) 00:58, 19 January 2008 (UTC)[reply]

How about an example where we're seeking strictly real solutions to a quadratic equation that also has "extraneous" imaginary solutions? Alternatively, maybe we create an example where we have an explicit given domain where we are only interested in positive solutions - but, there are also negative extraneous solutions. Tparameter (talk) 01:56, 19 January 2008 (UTC)[reply]

I agree that the example is rather transparent, but students learning algebra do stumble over such things. We want to keep the example simple. I don't know if there is an "official" definition of solutions being extraneous, but the way I understand the term they are solutions introduced by transforming the equation and not solutions arising by interpreting the equation in an extended domain.
One possible replacement is:
1/(x2+3x−2) = 1/(x2+x+2).
Take the inverse of both sides:
x2+3x−2 = x2+x+2.
Bring everything to one side:
2x−4 = 0.
So x = 2 – which is not a solution.
Another possible example:
x2+x+1 = 0.
Multiply by x:
x3+x2+x = 0.
Subtract the original equation:
x3−1 = 0.
So x = 1 – which is not a solution.
 --Lambiam 06:30, 19 January 2008 (UTC)[reply]
Let me define the term. An extraneous solution is a solution. That is first and foremost the most general definition. What kind of solution is it? It has the trait of being "extraneous", which we have to explain. So, I would say that an extraneous solution is a solution that is not applicable to a particular problem as it is defined. But remember, it is a solution. Tparameter (talk) 07:14, 19 January 2008 (UTC)[reply]