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There are only two situations that I am aware of that give rise to extraneous roots, namely, the “square both sides” situation (in order to eliminate a square root symbol), and the “half absolute value expansion” situation (in order to eliminate taking absolute value). An example of the former is $\sqrt{x} = x – 2$, and an example of the latter is $|2x – 1| = 3x + 6$. In the former case, by squaring both sides we get roots of $1$ and $4$, and inspection reveals that $1$ is extraneous. (Of course, squaring both sides is a special case of raising both sides to an positive even power.) In the latter case we expand the equation into the two equations $2x – 1 = 3x + 6$ and $2x – 1 = -(3x + 6)$, getting roots of $-1$ and $-7$, and inspection reveals that $-7$ is extraneous. Now, my question is: Is there any other situation besides these two that gives rise to extraneous roots? -Perhaps something involving trigonometry?

I asked this question some time ago in MO, where I got ground in the dirt like a wet french fry (as Joe Bob would say). So, I’m transferring the question here to MSE. :)

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Extraneous roots come up in log equations. – The Chaz 2.0 Aug 3 '11 at 22:46
Extraneous roots happen whenever you apply functions to both sides of an equation that aren't invertible. – Qiaochu Yuan Aug 3 '11 at 22:49
The two examples that you give can be considered as similar, since $|2x-1|=\sqrt{(2x-1)^2}$. – André Nicolas Aug 3 '11 at 22:55
@Andre: An excellent observation, thanks! – Mike Jones Aug 3 '11 at 23:25
@MikeJones One can fully avoid extraneous solutions by dealing with branches correctly. Take a radical problem, something along the lines $\sqrt x=x-2$, and solve using the quadratic formula, right? Wrong. The only possibly correct answer is simply the positive radical (the $\pm$ part) when you apply the quadratic formula, since all radicals in the original equation are defined as positive or vice versa. Weird things like that come up from time to time, as your problem attempts to have multiple branches all competing for the answer. But now, you can cross off answers without trying. – Simple Art Jun 16 at 14:46
up vote 15 down vote accepted

Suppose you have two expressions $e_1$ and $e_2$ and you know $$e_1 = e_2.$$
Then, if you apply a function to both sides, you have $$f(e_1) = f(e_2).$$ However, this logic in general does not reverse, unless the function $f$ is 1-1. This is the mechanism by which extraneous roots get introduced.

When you square both sides of an equation, you are destroying information about the signs of the two sides. Now, the equality will match if the two sides have the same absolute value. This process can, and often does, introduce spurious roots.

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@ncmahtsadist: OK, I'll go with that, upvoting, and accepting, your answer. Thanks! – Mike Jones Aug 3 '11 at 23:26
Ncmathsadist, can you be more specific as to how destroying signs will affect the number of solution? – Doeser Jan 28 '15 at 19:09
No square root can be equal to negative number. And square of positive is positive – Doeser Jan 28 '15 at 19:10

Extraneous solutions are often the result of omitting a constraint during the formulation or solution of a problem. For example, the correct rule for solving absolute value equations is

$$|x| = y \iff (y \ge 0) \text{ and } ((x = y) \text{ or } (x = -y)).$$

If we use this rule then extraneous solutions do not occur.

$$|2x - 1| = 3x + 6,$$ $$(3x + 6 \ge 0) \text{ and } (2x-1 = 3x+6 \text{ or } 2x-1 = -3x-6),$$ $$(x \ge -2) \text{ and } (x = -7 \text{ or } x = -1),$$ $$x = -1.$$

However, it is customary to omit the condition $y \ge 0$ and instead use the weaker rule $$|x| = y \implies ((x = y) \text{ or } (x = -y)).$$ This makes the writing simpler, but the price you pay is that you have to check for extraneous solutions at the end.

Extraneous solutions often arise from using a rule of the form $$x = y \implies f(x) = f(y).$$ Squaring both sides of an equation is an example of such a rule.

If $f$ is one-to-one, then the rule $$f(x) = f(y) \iff x = y$$ is valid, provided that $x$ and $y$ are both in the domain of $f$. Ignoring this condition can lead to extraneous solutions. The equation $\log(x-4) = \log(2x-6)$ provides an example.

Extraneous solutions can also result from ignoring physical constraints in applied problems (e.g. length and mass are positive quantities).

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Excellent! I've up-voted your answer. – Mike Jones Aug 4 '11 at 20:30

extraneous roots can occur when solving rational equations - this results when multiplying the equation to clear the fractions and cancelling a 0/0 term. you can also end up with extraneous roots with logarithmic equations because you will end up with x values that make the argument negative (not in the domain of log functions)

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