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)

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)


 * 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)


 * 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)


 * 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)


 * 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)

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


 * 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)


 * 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)


 * 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)


 * 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)


 * By "domain" I did not mean the domain of a function, but the domain of discourse, which in this case happens to be a domain in the sense of ring theory: the ring (and even field) of the real numbers. --Lambiam 03:14, 21 January 2008 (UTC)


 * I like the initial example, but I agree that the wording could use revision to make it a bit less condescending and magical. Keep in mind that this page is targeted at a high school algebra audience, who are less likely to inspect an equation for nonsolutions up front (but it can still be formal in tone). Dcoetzee 18:13, 31 January 2008 (UTC)

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)

No, this is not a sutiable example of an extraneous solution. Since x = -4 can satisfly the equation x2 - 3x + 5 = 0, only does not satisfly the domain that sets manually. Note that extraneous solution should focus on the naturally-formed domain but not on the manually-formed domain. Since you can set manually-formed domain at you own taste but naturally-formed domain cannot. naturally-formed domain should comes from the orginal equation. If the solution that does not satisfly the naturally-formed domain is also called extraneous solution, it should be trouble. Since this time all equation may be generate "extraneous solution" if you set an appropriate manually-formed domain. The term extraneous solution should not follow this approach.

If you search extraneous solution in Yahoo!, you can find many topics talking about extraneous solution. http://www.jcoffman.com/Algebra2/ch1_5.htm and http://www.mathpath.org/proof/argument.invalid.htm are two appropriate examples.Doraemonpaul (talk) 19:51, 18 February 2009 (UTC)

Neither 7 nor -4 satisfy the equation -- regardless of what our domains are. Just plug them into the equation. Do the math and you will see too. Also, I think it was great that you explained how a solution is not considered extraneous if it only does not satisfy the manually-formed domain (as opposed to the naturally-formed domain). The word is "satisfy," by the way -- not "satisfly." MusicHuman (talk) 14:52, 17 December 2022 (UTC)

My changes
I expanded on this idea a bit, starting with some very simple examples, including discussion of operations that remove solutions, and outlining a more general and (hopefully) more understandable categorization of what operations do what to the solution set. Feel free to revise. Dcoetzee 20:31, 31 January 2008 (UTC)

Possible errors
I see a couple of things in this article that don't seem right. Firstly, with the example $$\sqrt{x}=-1$$, the article says x=1 is not a solution. That means $$\surd$$ is being used to mean positive square root (as is common), but in that case the equation has no solutions by definition, so is a rather poor example. Secondly, at the end it says multiplication is surjective by not injective. It's quite clearly not surjective since it's image is just {0}. Am I missing something here, or should I just fix it? --Tango (talk) 23:24, 1 March 2008 (UTC)


 * I've removed the last section because it is mostly incorrect, not easy to fix, and essentially OR. I wish the article had not been expanded to cover missing solutions, which makes everything a lot more complicated, and as far as I know is not a common term to describe errors that may arise in solving high-school algebra problems. --Lambiam 15:45, 2 March 2008 (UTC)


 * Removed the incorrect example $$\sqrt{x}=-1$$, where the article stated x=1 is not a solution: -1 is most certainly one of the second roots of unity. The missed solution x=1 stems from the fact that roots cannot be treated as "unknowns" that have an explicit value. Taking the nth root of x is precisely the statement, "at least one of the n roots of x satisfies this equation" -- the "positive square root" operator does not exist for just this reason. Sorry, your calculator lies to you :-) 68.42.7.184 (talk) 07:32, 23 April 2009 (UTC)

Extraneous solutions in applied problems
There is another type of extraneous solution, that is not mentioned in the article. A solution to an equation arising from an applied problem may be considered extraneous if it is not physically meaningful. A negative length could be an example of this. See http://mathcentral.uregina.ca/QQ/database/QQ.09.02/paul2.html 66.41.7.193 (talk) 05:00, 29 March 2008 (UTC)

Example with the imaginary unit, i
Is this a useful example? It got me confused for a while just now:

We know that $$\frac{1}{i}=-i$$. However,

$$\frac{1}{i}=i^{-1}=((-1)^{\frac{1}{2}})^{-1}=\sqrt{(-1)^{-1}}=\pm i$$

But only $$-i$$ satisfies the original equation, $$+i$$ is an extraneous solution.

--MTres19 (talk) 19:38, 29 April 2020 (UTC)