Extraneous Solutions – Part 1 of 3?

Disclaimer

Within my small inner circle of math teachers, the mystery of extraneous solutions seems to be the issue of the year. I have so much to say on this topic (algebraic, logical, pedagogical, historical, linguistic) that I don’t really know where to begin. My only disclaimer is that I’m not really sure if this topic is all that important.

Solving an Equation with a Radical Expression

Consider the following equation:

(1) $2\sqrt{x+8} +5 = 11$

One hardly needs algebra skills or prior knowledge to solve this, but prior experience suggests trying to isolate $x$ .

(2) $2\sqrt{x+8} = 6$ (we subtract 5 from both sides)

(3) $\sqrt{x+8} = 3$ (we divide both sides by 2)

Now, if the square root of something is 3, then that something must be 9, so it immediately follows that

(4) $x+8 = 9$

(5) $x = 1$ (we subtract 8 from both sides)

Squaring Both Sides

In my transition from (3) to (4), I used a bit of reasoning. Some conversational common sense told me that “if the square root of something is 3, then that something must be 9”. But that logic is usually just reduced to an algebraic procedure: “squaring both sides”. If we square both sides of equation (3), we get equation (4).

On the one hand, this seems like a natural move. Since the meaning of $\sqrt{a}$ is “the (positive) quantity which when squared is $a$ “, the expression $\sqrt{a}$ is practically begging us to square it. Only then can we recover what lies inside. A quantity “which when squared is $a$ ” is like a genie “which when summoned will grant three wishes”. In both cases you know exactly what to do next.

Unfortunately, squaring both sides of an equation is problematic. If $a = b$ is true, then $a^2 = b^2$ is also true. But the converse does not hold. If $a^2 = b^2$ , we cannot conclude that $a = b$ , because opposites have the same square.

This leads to problems when solving an equation if one squares both sides indiscriminately.

A Silly Equation Leads to Extraneous Solutions

Consider the equation,

(6) $x = 4$

This is an equation with one free variable. It’s a statement, but it’s a statement whose truth is impossible to determine. So it’s not quite a proposition. Logicians would call it a predicate. Linguistically, it’s comparable to a sentence with an unresolved anaphor. If someone begins a conversation with the sentence “He is 4 years old”, then without context we can’t process it. Depending on who “he” refers to, the sentence may be true or false. The goal of solving an equation is to find the solution set, the set of all values for the free variable(s) which make the sentence true.

Equation (6) is only true if $x$ has value 4. So the solution set is $\left\{4 \right\}$ . But if we square both sides for some reason…

(7) $x^2 = 16$ has solution set $\left\{4, -4\right\}$

We began with $x = 4$ , “did some algebra”, and ended up with $x^2 = 16$ . By inspection, $-4$ is a solution to $x^2 = 16$ , but not to the original equation which we were solving, so we call $-4$ an “extraneous solution”. [Extraneous – irrelevant or unrelated to the subject being dealt with]

Note that the appearance of the extraneous solution in the algebra of (6)-(7) did not involve the square root operation at all. But this example was also a bit silly because no one would square both sides when presented with equation (6), so let’s look at a slightly less silly example.

Another Radical Equation

(8) $2\sqrt{x+8} + 5 = -1$

(9) $2\sqrt{x+8} = -6$

(10) $\sqrt{x+8} = -3$

People paying attention might stop here and conclude (correctly) that (10) has no solutions, since the square root of a number can not be negative. Closer inspection of the logic of the algebraic operations in (8)-(10) enables us to conclude that the original equation (8) has no solutions either. Since $a = b \iff a - 5 = b -5$ , any solution to (8) will also be a solution to (9) and vice versa. Since $a = b \iff a/2 = b/2$ , any solution to (9) will also be a solution to (10) and vice versa. So equations (8), (9), and (10) are all “equivalent” in the sense that they have the same solution set.

But what if the equation solver does not notice this fact about (10) and decides to square both sides to get at that information hidden inside the square root?

(11) $x+8 = 9$

(12) $x = 1$

Again we have an extraneous solution. $x = 1$ is a solution to (12), but not to the original equation (8). Where did everything go wrong? By the previous logic, (8), (9), and (10) are all equivalent. (11) and (12) are also equivalent. So the extraneous solution somehow arose in the transition from (10) to (11), by squaring both sides.

So unlike subtracting 5 from both sides or dividing both sides by 2, squaring both sides is not an equivalence-preserving operation. But we tolerate this operation because the implication goes in the direction that matters. If $a = b$ , then $a^2 = b^2$ , so if $a$ and $b$ are expressions containing a free variable $x$ , any value of $x$ that makes $a = b$ true will also make $a^2 = b^2$ true.

In other words, squaring both sides can only enlarge the solution set. So if one is vigilant when squaring both sides to the possible creation of extraneous solutions, and is willing to test solutions to the terminal equation back into the original equation, the process of squaring both sides is innocent and unproblematic.

Those Who are Still Not Satisfied

Still there are some who are not satisfied with this explanation: “Why does this happen? What is really going on? Where do the extraneous solutions come from? What do they mean?”

One source of the problem is the square root operation itself. $\sqrt{a}$ is, by the conventional definition, the positive quantity which when squared is $a$ . The reason that we have to stress the positive quantity is that there are always two real numbers that when squared equal any given positive real number. There are a few slightly different ways of making this same point. The operation of squaring a number erases the evidence of whether that number was positive or negative, so information is lost and we are not able to reverse the squaring process.

We can also phrase the phenomenon in the language of functions. Since squaring is a common and useful mathematical practice, information will often come to us squared and we’ll need an un-squaring process to unpack that information. $f(x) = x^2$ , for all the reasons just mentioned, is not a one-to-one function, so strictly speaking, it is not invertible. But un-squaring is too important, so we persevere. As with all non-one-to-one functions, we first restrict the domain of $f(x) = x^2$ to $[0, \infty)$ to make it one-to-one. This inverse, $f^{-1}(x) = \sqrt{x}$ thus has a positive range and so the convention that $\sqrt{a} \geq 0$ is born. So every use of the square root symbol comes with the proviso that we mean the positive root, not the negative root. We inevitably lose track of this information when squaring both sides.

[Note: Students can easily lose track of these conventions. After a lot of practice solving quadratic equations, moving from $x^2 = 9$ effortlessly to $x = \pm 3$ , students will often start to report that $\sqrt{9} = \pm 3$ .]

The convention that we choose the positive root is totally arbitrary. In a world in which we restricted the domain of $f(x) = x^2$ to $(-\infty, 0]$ before inverting, $\sqrt{9}$ would be $-3$ . In that world, $x = 1$ is a perfectly good solution to $2\sqrt{x+8} + 5 = -1$ , not extraneous at all.

A Trigonometric Equation which Yields an Extraneous Solution

For parallelism, consider the (somewhat artificial) equation:

(13) $\arccos(2x-1) = \frac{4\pi}{3}$

Like in (10), careful and observant solvers might notice that the range of the $\arccos(x)$ function is $[0, \pi]$ and correctly conclude that the equation has no solutions. But there seems to be a lot going on inside that $\arccos$ expression, so many will rush ahead and try to unpack it by “cosineing”. Indeed, since $a=b \Rightarrow \cos(a) = \cos(b)$ , this seems innocent.

(14) $2x - 1 = -\frac{1}{2}$

(15) $2x = \frac{1}{2}$

(16) $x = \frac{1}{4}$

But $x = \frac{1}{4}$ is an extraneous solution since $\arccos(-\frac{1}{2}) = \frac{2\pi}{3}$ not $\frac{4\pi}{3}$ .

The explanation for this extraneous solution will be similar to the logic we used above. If $a = b$ , then $\cos(a) = \cos(b)$ , so if $a$ and $b$ are expressions containing a free variable $x$ , any value of $x$ that makes $a = b$ true will also make $\cos(a) = \cos(b)$ true. So we will not lose any solutions by “taking the cosine of both sides”. But as the cosine function is not one-to-one, $\cos(a) = \cos(b)$ does not imply that $a = b$ . So taking the cosine of both sides, just like squaring both sides, can enlarge the solution set.

The above paragraph explains why extraneous solutions could appear in the solution of (13), but maybe not why they do appear. For that, we again must look to the presence of the $\arccos$ function. Since $\cos$ is not one-to-one, we had to arbitrarily restrict its domain to $[0, \pi]$ prior to inverting. So every use of the $\arccos$ symbol comes with its own proviso that we are referring to a number in a particular interval of values. In a world in which we had restricted the domain of $\cos$ to $[\pi, 2\pi]$ prior to inverting, $x = \frac{1}{4}$ would be a perfectly good solution to $\arccos(2x-1) = \frac{4\pi}{3}$ , not extraneous at all.

The above examples seem to suggest that one can avoid dealing with extraneous solutions by carefully examining one’s equations at each step. But in practice, this really isn’t possible. I saved the fun examples for the end, but as this post is already way way too long, they will have to wait for a bit later.

-Will Rose

Thanks

Thanks to John Chase for letting me guest post on his blog. Thanks to James Key for encouraging me again and again to think about extraneous solutions.

9 thoughts on “Extraneous Solutions – Part 1 of 3?”

howardat58 on January 21, 2017 at 6:30 am said:

“One source of the problem is the square root operation itself. \sqrt{a} is, by definition, the positive quantity which when squared is a.”

“By definition…” is mathematically too restrictive. It should be abandoned. The square root of x has two values.
Try the square root of i, or more precisely i itself. Is i positive or negative?
What about the cube root of 1?

Reply ↓
- Mr. Chase on January 21, 2017 at 9:20 am said:
  
  Howard, the bigger issue is whether non-invertible functions may be applied to both sides of an equation. I think all algebraic practitioners agree that such functions can be applied to both sides of an equation. This is the bigger idea Will is trying to make, I think.
  
  Depending on the context, we mean different things by the square root function. It’s perfectly acceptable in the context of the real numbers to speak of the principal square root. The moment you entertain the idea of multi-valued functions, you change the discussion entirely.
  
  If we DO consider multi-valued functions, how do you think the discussion changes? That is, how do extraneous solutions arise in this more complicated environment, or do they at all? I have some ideas, but I haven’t thought about it very much yet. I’d love to hear what you think!
  
  Reply ↓
Mr. Chase on January 21, 2017 at 8:41 am said:

Brilliant guest post, Will.

I’m a huge fan of explaining extraneous solutions in terms of non-invertible functions having been applied to both sides of an equation. Only in the case that a function is one-to-one can we ensure that the next equation we produce has the same solution set (“algebraically equivalent”). You said this very nicely and gave good examples.

I made the claim in my last post that the worldwide mathematical community agrees that the way we “do algebra” assumes forward implications. That is, we allow users of algebra to apply any function to both sides of an equation (not necessarily just one-to-one functions) because this ensures that the new equation follows from the first equation, IF the first equation holds.

This is pictured nicely by the “equation balance” physical model that often makes an appearance in prealgebra classrooms. As we do algebra, our only requirement is that we keep the balance.

I’m looking forward to the next installments in your series!

Reply ↓
- wrose31 on January 21, 2017 at 10:19 pm said:
  
  I don’t think that it totally makes sense when solving an equation to say that one equation “follows from” another. I know what you mean, but I think it’s more complicated than that. Technically these are not propositions which are either true or false and so it doesn’t completely make sense to say that one follows from another. The equations contain a free variable, so they’re technically predicates.
  
  We’re not manipulating true propositions in a truth-preserving way to produce new true propositions. When trying to solve an equation, we are seeking the solution set, so we’re manipulating the predicate to produce a new predicate with the same or possibly larger solution set.
  
  It’s correct to say “Will is happy” follows from “Will is eating a cookie” because the truth of the latter implies the truth of the former. But it’s awkward/misleading to say that “He is happy” follows from “He is eating a cookie”. Yes, it’s true that for any particular referent of “he”, the truth of the latter implies the truth of the former, but the relationship between the solution sets of the two sentences can not adequately be captured with “follows from” language, in my opinion.
  
  Reply ↓
  - Mr. Chase on January 21, 2017 at 10:23 pm said:
    
    I guess when you write Equation A followed by Equation B, I interpret it as “if Equation A holds for some x, then Equation B holds for that x as well.”
    
    Is that fair?
howardat58 on January 22, 2017 at 8:16 pm said:

A simple explanation is
square root of x-squared = x

The principal value of square root of x-squared is mod x
and this creates algebraic difficulties.

Reply ↓
- Tino Pham on January 26, 2017 at 1:26 am said:
  
  The sqrt(x^2) is the absolute value of x, if x is being restricted to real numbers. Though, I agree with you, standard definition is too restrictive. By standard definition, the square root symbol is the unique nonnegative square root of a nonnegative real number.
  
  I feel like the problem with noninvertible processes (squaring and “square rooting”) is If we were to square a number and try to “undo” the process by taking the square root, we cannot determine the unique original number. Since x^2 is not a one-to-one function, its inverse isn’t a function where its composition of x^2 and the “square root” of x gives us x in the real numbers.
  
  By arbitrarily restricting the domain of a one-to-one function, we get even weirder “extraneous solutions” such as in inequalities like x^2 > 9. If we were to “square root” both sides, we get x > 3 or x > -3, which isn’t true at all! However, since sqrt(x^2) is the absolute value of x, then the inequality becomes | x| > 3, then x >3 or x< -3. Hence, The
  
  Reply ↓
  - Mr. Chase on January 31, 2017 at 9:50 pm said:
    
    …dying to know the end of your sentence, Tino! 🙂
- Mr. Chase on January 31, 2017 at 9:54 pm said:
  
  Yeah, I mentioned this in reply to your earlier comment. It’s easier to just restrict our attention to real numbers and consider the square root function as returning only the positive root (principal value).
  
  I wrote a little article on complex roots here, which might be relevant. Also relevant might be this nice section of the Wikipedia article on Exponentiation. It gives some interesting failures of our normal exponentiation rules when applied to complex numbers. It’s all about crossing that branch cut! Makes a mess of everything.
  
  Reply ↓

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Extraneous Solutions – Part 1 of 3?

Disclaimer

Solving an Equation with a Radical Expression

Squaring Both Sides

A Silly Equation Leads to Extraneous Solutions

Another Radical Equation

Those Who are Still Not Satisfied

A Trigonometric Equation which Yields an Extraneous Solution

Thanks

9 thoughts on “Extraneous Solutions – Part 1 of 3?”

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Disclaimer

Solving an Equation with a Radical Expression

Squaring Both Sides

A Silly Equation Leads to Extraneous Solutions

Another Radical Equation

Those Who are Still Not Satisfied

A Trigonometric Equation which Yields an Extraneous Solution

Thanks

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9 thoughts on “Extraneous Solutions – Part 1 of 3?”

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