Talk:Extraneous and missing solutions

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 try to 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]

Unless of course, it is a misnomer. I have just been instructed by a colleague that the term is a misnomer, and actually does not describe solutions at all. To avoid original research, I'm not touching this article, I might make it much worse. Tparameter (talk) 07:40, 19 January 2008 (UTC)[reply]

Of course every value V is a solution (namely of the equation x = V). The meaning of a term is what people who use the term in general mean when they use it. That is something that cannot always be determined by juxtaposing the usual meanings of components of the term in question. A Welsh rabbit is not a rabbit that is Welsh and the division algorithm is not an algorithm. Based on how the term "extraneous solution" is used, it is clear that the terms refers to any solution V of some equation E' such that E', viewed as a proposition, is a valid consequence of an original equation E, but which value V, however, is not a solution of E. Moreover, V has to be found as a solution to E' in the process of attempting to solve E. It follows that the solutions of E are a subset of those of E'. The solution set of E' consists of the solution set of E plus, possibly, some other solutions, called extraneous solutions. --Lambiam 15:43, 19 January 2008 (UTC)[reply]

Lambiam, on your first example, sorry, why isn't 2 a solution? Tparameter (talk) 16:08, 19 January 2008 (UTC)[reply]

Oops, I swapped the two sides during editing but made an error; it should have read

1/(x²+3x+2) = 1/(x²+x−2),

resulting in an extraneous solution of x = −2. --Lambiam 16:46, 19 January 2008 (UTC)[reply]

So, am I clear then that one form of extraneous solutions are simply "solutions" not in the domain of x? Tparameter (talk) 17:49, 19 January 2008 (UTC)[reply]

I don't think so; as I said before interpreting an equation in another domain does not amount (as I see it) to transforming it into another equation. In the context of elementary algebra in which this terminology is used the students probably haven't even been introduced to the notion of solving in other domains than the reals. --Lambiam 00:56, 20 January 2008 (UTC)[reply]

Isn't it true that in your example above that x = -2 is not in the domain of x? Tparameter (talk) 14:11, 20 January 2008 (UTC)[reply]

Is this an example of an extraneous solution?

The domain is given as the set of all ${x:x\in {\mathbb {R}},x>0}$ . Given the domain, and given

x² - 3x + 5 = 0

Find x. Using the quadratic formula, it is easy to find the solutions

x = 7 and x = -4

where x = 7 is the solution to the problem, and x = -4 is an extraneous solution because it is not pertinent to the problem. Tparameter (talk) 17:46, 19 January 2008 (UTC)[reply]