The other day a student I was helping face to face asked how she can know when to check for extraneous solutions of an equation. I gave her a quick version of my standard answer, and the light went on! Today I want to share these thoughts here, because they are very important in several ways. In searching for references, I found that I have been saying essentially the same thing for a long time. Some of it, I don’t think I’ve ever seen a textbook state, though I’m sure I didn’t invent it myself.

## What causes extraneous solutions?

Here is a question that will serve as a good overview. Sarah in 2013 asked,

Extraneous Routes I know that radical, log, absolute value, and fractional operations sometimes introduce extraneous roots. When else do I need to check for them? Is it only by an equation?

In case you are not familiar with the concept, an extraneous solution (also called an extraneous root) is a solution you get in the process of solving, that turns out not to be a solution of the original equation. It is *not* actually a solution, and is not inherent in the problem itself, but is introduced by what you do.

I sometimes illustrate the idea by imagining that we work in a lab analyzing blood samples. Suppose we add some reagent to the sample in order to determine, say, whether there is any arsenic in the blood, and then (still using the same sample) we do a test for some other class of poisons – but the reagent we used for the first test is one of them. Then we will find that chemical in our sample, not because it was originally there (though it may have been), but because we put it there! We need to know that we* introduced it* into the sample, and either ignore it or do some other test to see if it should be included on our report. In the same way, we need to pay attention to what we have done that may introduce an extraneous solution, so we can check for it in the end.

Sarah’s question was a broad one, so I gave a quick survey of the concept, emphasizing that it is not primarily the *type of equation* that requires the check (though that is a good clue), but *what you do* in solving it:

You have to check for extraneous roots whenever -- in the process of solving -- you have done something that is not guaranteed to produce an equivalent equation. I wouldn't want to claim to give a complete list; any time you use a technique you haven't used before, you should determine for yourself whether it falls in this category. But the most familiar of these aremultiplying by an expression containing the variable, which you do in solvingrational equations, and which might result in unintentionally multiplying by zero;squaring or raising to another even power, which you do in solvingradical equationsor equations withfractional exponents, and which loses information about signs; andsimplifying logarithmic expressions, which can change the domain of an expression. Certain things you do in solvingabsolute valuescan also do this, though in many cases this can be handled by paying close attention to conditions rather than just checking at the end. What matters most is not the kind of equation you are solving so much as the things you do in solving it. For example, sometimes people solve absolute value equations by squaring -- that falls under one category I listed. Other people will do something different that does not carry any risk.

Let’s look at each of these main types, to fill in some details.

### Multiplying by the variable:

Suppose we start with the equation \(x = 3\), and then (just for fun) multiply both sides by *x*. What do we get? \(x^2 = 3x\). If you solve this new equation (be careful!), you find that it has two solutions, 3 and 0. The original equation had only one solution, 3 (obviously). So the new equation is not equivalent to the original; the new one has an extra solution, 0. Why? Because when you multiply by 0, both sides become 0, and what may have been a false equation is now a true one. So any value of *x* for which the multiplier is zero will look like a solution, whether or not it was.

Of course, we wouldn’t multiply that equation by *x* to solve it; but we *would* do this to solve a **rational equation** like \(\frac{1}{x} + 2 = \frac{x + 1}{x}\); after multiplying by *x*, we have \(1 + 2x = x + 1\), and we find that the solution is \(x = 0\). But if *x* is zero, then we multiplied by zero, so of course that will appear to be a solution! In fact, if we check \(x = 0\) in the original equation, we find that both sides are undefined, which means that 0 is not a solution.

Note that there is something extra happening here: When we multiplied by *x*, we also *changed the domain* of the equation, by eliminating the denominator. In fact, this is what we really need to check for: If our claimed solution is not in the domain of the original equation, it is extraneous, and must be ignored.

### Squaring each side:

Suppose we start with the same equation \(x = 3\), and then square both sides. This time we get \(x^2 = 9\), which has two solutions, 3 and -3. Again, the new equation is not equivalent to the original; we introduced the extra “solution” -3. Why? Because when you square, you lose information about signs. If we had started with \(x = -3\), we would have ended up with the same equation. So when we solve the squared equation, we are actually solving *both* original equations at once; our “solutions” may be solutions of one, or the other, or both, and we don’t know which until we check.

This typically happens when we are solving a **radical equation**. As an example, we might solve \(\sqrt{x + 6} = x\) by squaring, which yields \(x + 6 = x^2\). The solutions of this new equation are 3 and -2. The first of these, 3, is in fact a solution of the original equation: \(\sqrt{3 + 6} = 3\). But -2 is *not* a solution: \(\sqrt{-2 + 6} = 2\), not -2. What happened? We actually solved the negated equation, \(-\sqrt{x + 6} = x\), of which -2 *is* the solution!

There is more going on here, too. Sometimes (though not usually), the equation is undefined because the radicand becomes negative; an example of this is \(\sqrt{2x + 1} = \sqrt{x}\). Squaring yields \(2x + 1 = x\), whose solution is \(x = -1\). When we check this, we find that -1 is not in the domain of the equation, so that rather than being a solution of the negated equation, it is not a solution of either. The usual problem, as in the first example above, can also be described as a *range* issue: because the radical symbol represents only the *positive* square root, it is this restriction on the range that caused our -2 not to be a solution.

Here is an example from 2001 where a student discovered this for himself:

Extraneous Roots

### Simplifying a logarithmic or other expression:

Domain issues are the usual culprit in logarithmic equations. Here is an example: \(\log(x) + \log(x+3)= 1\). If we simplify (condense) the left side, we get \(\log(x^2 + 3x) = 1\) and then \(x^2 + 3x = 10\), whose solution is \(x = 2, -5\). But because of the domain of the log, the negative solution is extraneous: \(\log(2) + \log(2+3)= 1\) is true, but \(\log(-5) + \log(-5+3)= 1\) is not. The issue here is that condensing a logarithmic expression (or some other types of expression) can change the domain. For more on this, see

Are Properties of Logarithms Missing Something?

## How can you recognize extraneous solutions?

Hidden in the details above is an important fact: each kind of extraneous solution has its own specific test. Students too often miss the fact that when they check a solution, it can fail for two very different reasons: It may fail because it is an extraneous solution that we introduced by our work (in which case the check is an essential part of the work itself); or it may fail just because it is wrong, a result of a mistake in our work. If it is **extraneous**, we can just ignore it (and say there is no solution, if we found no non-extraneous solutions); if it is **erroneous**, we have to go back and fix our work.

So when I teach about this, I always show how to distinguish an extraneous solution from an erroneous solution.

### Rational equations:

Here is a question from 2012:

Extraneous Roots Checked Less Tediously Given (x - 4)/x + x/3 = 6 I solved this for x and found two roots: x = [15 - SQRT(273)]/2 x = [15 + SQRT(273)]/2 Can anybody help me check whether one of these roots is extraneous or not? Plugging in these values of x and satisfying the left hand side and right hand side would take too long. For simple values of x, like 2 and 3, it's easier to find the extraneous roots. But values of x that have two parts and contain square roots, as in my case, cause difficulty in finding the extraneous roots. Other than plugging in values, is there a simple way of checking for extraneous roots?

Muhammad knew the need to check the solutions, but the check here seemed too hard, since the solutions were ugly expressions. I gave him two pieces of advice that can simplify the checking process. First, since any extraneous solution to a rational equation will fail by making terms undefined, he only needs to check that:

In this kind of equation, the source of extraneous roots is multiplication by x, which can introduce an extraneous root if x = 0 turns out to be a root of the new equation, because multiplication by 0 does not produce an equivalent equation. To put it another way, an extraneous root will prove to be extraneous only by making a denominator of the original equation zero. So all you need to do to test for an extraneous root is to make sure each solution is in the domain of the equation -- that is, none makes the denominator zero. In this case, it is clear that they don't.

Second, checking is also needed in order to make sure you didn’t make a mistake; but there, the accuracy of a calculator is good enough; you don’t need exact verification unless you are told to:

The first root, for example, is approximately -0.76135582. I would store that value in my calculator to avoid having to retype it three times, and use it in the equation: (x - 4)/x + x/3 = (-0.76135582-4)/-0.76135582 + -0.76135582/3 = 6.25378527 + -0.25378527 = 6 If this had come out to 5.99999999, I would consider it verified!

I also gave a warning that many students need to hear: Once they have learned about extraneous solutions, they often start writing “no solution” whenever a check fails, even if it is a linear equation that can’t possibly be extraneous! Their knowledge of extraneous solutions seems to have displaced what they formerly knew, that they themselves are the most common cause of failed checks. Only ignore failed solutions that you *know* are extraneous.

### Radical equations:

Here is a question from 2017 about recognizing extraneous solutions caused by squaring:

One Variable in Two Radicals Solve for x: sqrt(2x - 5) - sqrt(x - 2) = 2 I tried squaring each term individually and then squaring the 2, but my roots are not the roots in the solution. ...

Here Phinah had several different issues, so I went through the whole process of solving; but at the end (before being asked) I pointed out how checking works here:

I'll mention one extra thing: when you are checking your answer, you can recognize an extraneous root (as opposed to an error due to, for example, an arithmetic mistake) if it satisfies an equation obtained by changing the sign of a radical. In this case, when you check x = 3, you get sqrt(2x - 5) - sqrt(x - 2) = 2 sqrt(2*3 - 5) - sqrt(3 - 2) = 2 sqrt(1) - sqrt(1) = 2 This is false. But it becomes true if you change a sign to sqrt(1) + sqrt(1) = 2 So this is actually a solution of sqrt(2x - 5) + sqrt(x - 2) = 2 This is indistinguishable from the given equation after squaring. That is why the extraneous solution arises.

This is typical: If the check fails, see if changing the sign on one or both radicals would make it correct. If so, then the solution is extraneous and you can quietly cross it out; but if not (if you got, say, \(\sqrt{3} – \sqrt{2} = 2\), which is simply false), then you have to find the error, because your work was wrong.

## Some related issues

There are some other things worth discussing that are related to extraneous solutions, but I will just provide links:

Here I offer an alternative (which I had never thought of previously) that simplifies the check in some cases:

Avoiding the Final Step of Checking for Extraneous Solutions

Here I discussed the opposite issue, where you can lose a valid solution rather than find an invalid one:

Root Propagation and Loss

And here I discussed similar issues in trigonometric equations:

Extraneous Roots Introduced Trigonometrically Algebra for Equivalence

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AnonymousSay I have

Sqrt(x-3)=4

I solve the equation and get x=19. When I plug in the equation I get…

Sqrt(16)=4

Why don’t I have to do…

4=4 AND -4=4

Dave PetersonIn a sense, it’s the same reason you can get extraneous solutions to radical equations.

The radical symbol, \(\sqrt{16}\), represents only the positive root; so you only have 4, not -4, on the left-hand side. We define the radical as representing only one value so that it is a

function, which makes algebra easier.If you were solving the equation \(\sqrt{x-3}=-4\), you would get the same solution, \(x=19\), but that would be extraneous because \(\sqrt{16}=4\), not \(-4\).

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