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==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? [[User:Tparameter|Tparameter]] ([[User talk:Tparameter|talk]]) 00:58, 19 January 2008 (UTC)
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== Is this an example of an extraneous solution? ==
The ___domain is given as the set of all <math>{x : x \in \
Given the ___domain, and given
:x<sup>2</sup> - 3x + 5 = 0
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No, this is not a sutiable example of an extraneous solution. Since x = -4 can satisfly the equation x<sup>2</sup> - 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
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." [[User:MusicHuman|MusicHuman]] ([[User talk:MusicHuman|talk]]) 14:52, 17 December 2022 (UTC)
== My changes ==
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:I've removed the last section because it is mostly incorrect, not easy to fix, and essentially [[WP:OR|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. --[[User talk:Lambiam|Lambiam]] 15:45, 2 March 2008 (UTC)
:Removed the incorrect example <math>\sqrt{x}=-1</math>, 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 n<sup>th</sup> 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 :-) [[Special:Contributions/68.42.7.184|68.42.7.184]] ([[User talk: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 [[Special:Contributions/66.41.7.193|66.41.7.193]] ([[User talk: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 <math>\frac{1}{i}=-i</math>. However,
<math>\frac{1}{i}=i^{-1}=((-1)^{\frac{1}{2}})^{-1}=\sqrt{(-1)^{-1}}=\pm i</math>
But only <math>-i</math> satisfies the original equation, <math>+i</math> is an extraneous solution.
--[[User:MTres19|MTres19]] ([[User talk:MTres19|talk]]) 19:38, 29 April 2020 (UTC)
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