When do we get extraneous roots?

Question

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. :)

Answer 1

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.

Answer 2

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).