Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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. :)

share|improve this question
    
Extraneous roots come up in log equations. –  The Chaz 2.0 Aug 3 '11 at 22:46
5  
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
1  
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

3 Answers 3

up vote 10 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.

share|improve this answer
1  
@ncmahtsadist: OK, I'll go with that, upvoting, and accepting, your answer. Thanks! –  Mike Jones Aug 3 '11 at 23:26

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).

share|improve this answer
1  
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)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.