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. :)

edit (1.Jan.2017): In general, in mathematics and the real world, you get extraneous roots any time you are initially presented with (via some mechanical / automated / canonical process) a superset of the set that you want, and the sifting out of that set from the superset is left to you, for example, when panning for gold, or when reading the owner’s manual for your vehicle: “This owner’s manual covers all models of your vehicle. You may find descriptions of equipment and features that are not on your particular model.”

  • $\begingroup$ Extraneous roots come up in log equations. $\endgroup$ Aug 3, 2011 at 22:46
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    $\begingroup$ Extraneous roots happen whenever you apply functions to both sides of an equation that aren't invertible. $\endgroup$ Aug 3, 2011 at 22:49
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    $\begingroup$ The two examples that you give can be considered as similar, since $|2x-1|=\sqrt{(2x-1)^2}$. $\endgroup$ Aug 3, 2011 at 22:55
  • $\begingroup$ @Andre: An excellent observation, thanks! $\endgroup$
    – Mike Jones
    Aug 3, 2011 at 23:25
  • $\begingroup$ @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. $\endgroup$ Jun 16, 2016 at 14:46

5 Answers 5


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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    $\begingroup$ @ncmahtsadist: OK, I'll go with that, upvoting, and accepting, your answer. Thanks! $\endgroup$
    – Mike Jones
    Aug 3, 2011 at 23:26
  • $\begingroup$ Ncmathsadist, can you be more specific as to how destroying signs will affect the number of solution? $\endgroup$ Jan 28, 2015 at 19:09
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    $\begingroup$ No square root can be equal to negative number. And square of positive is positive $\endgroup$ Jan 28, 2015 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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    $\begingroup$ Excellent! I've up-voted your answer. $\endgroup$
    – Mike Jones
    Aug 4, 2011 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)


Here is an unusual extraneous root, based on a video by Pesh Talwalker, $3=0$, but slightly different. Consider $x^2-x-1=0$. This has the roots $(1\pm\sqrt(5))/2$ or $\phi$ and $-1/\phi$. $0$ is not a root. So divide both sides by $x$:

$x-1-1/x=0$. (*)

Now $x^2 = x+1$ so that

$x-1 = x+1-2 = x^2-2$

Substitute in (*) to get


Multiply both sides by $x$ to get


This equation has $\phi$ and $-1/\phi$ as roots all right, but it also has the extraneous root $-1$.

  • $\begingroup$ Notice that, when you substitute for x-1, you're applying a function to only 1 occurrence of x in equation (*), and in only one side of this equation. So, even though the function you're applying is reversible, the equation derivation isn't. You would get the same result if you substituted 1 + 1/x for x in the quadratic equation $x^2 - x - 1$. $\endgroup$
    – Papa Smurf
    Nov 18, 2021 at 19:29
  • $\begingroup$ I discussed this phenomenon in this answer. $\endgroup$
    – ryang
    Jun 22, 2022 at 21:06

Another less obvious way that extraneous roots may be introduced is through, lacking a better word, "careless commingling".

Suppose that an equation in $x$ of the form $f(x,a,b)=0$ is known to have $2$ solutions, and the problem asks to find $a,b$ so that the two solutions are $a$ and $b$. One natural approach would be to write the following system and solve it.

$$ \begin{cases} \begin{align} f(a,a,b) &= 0 \\ f(b,a,b) &= 0 \end{align} \end{cases} $$

The less obvious part is that the solutions of the system do not necessarily satisfy the original problem, in other words solving it this way potentially introduces extraneous solutions. Reason is that the problem assumes $a,b$ to be the roots of the equation, with emphasis on the plural of "roots". But the system of equations loses that distinction, and allows for $a$ and $b$ to be one and the same root.

Below is an example of this happening, based on an actual question.

Find $a,b$ knowing that the roots of the equation $x^2 - \lambda a x + b=0$ are $a$ and $b$.

Using Vieta's relations, the solution set is easily found to be $(a,b) \in \{\,(0,0), (1, \lambda-1)\,\}$.

But the system $ \;\begin{cases} \begin{align} a^2 - \lambda a^2 + b &= 0 \\ b^2 - \lambda a b + b &= 0 \end{align} \end{cases} \; $ has solutions $\,(a,b) \in \left\{\,(0,0), (1, \lambda-1), \left(\frac{1}{\lambda-1},\frac{1}{\lambda-1}\right)\right\}\,$ when $\lambda \ne 1$. The extraneous solution $\left(\frac{1}{\lambda-1},\frac{1}{\lambda-1}\right)$ was introduced by "careless commingling", and it satisfies the system but not the original problem.

FWIW not all solutions where $a=b$ are necessarily extraneous, for example $(0,0)$ is a valid solution with $a=b$, and so is $(1,\lambda-1),$ when $\lambda=2$. The only safe recourse in this case is to verify each solution individually (unless the case $a=b$ is handled completely separately).


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