Extraneous Solutions – Part 2 of 3

Solving an Equation as a Sequence of Equation Replacement Operations

Part 1 was so long because I wanted to be extremely thorough and to present things to an audience that perhaps hadn’t thought much about the logic of equation solving at all. Since we’re now all experts, perhaps it’s worth it to summarize everything very succinctly.

Given an equation in one free variable, we want to find the solution set. To do this, we replace that equation with an equivalent equation whose solution set is more obvious.

(1) 8x - 5 = 5x + 1

(2) 8x = 5x + 6

(3) 3x = 6

(4) x = 2

If in the transition from (1)-(2), from (2)-(3), and from (3)-(4) we are careful to replace each equation with an equivalent equation, then by the transitivity of equivalence, the original equation and terminal equation are guaranteed to be equivalent. Since the solution set of the terminal equation is obvious, we know the solution set of the original equation, as well. Thus solving an equation requires establishing that certain equation replacement operations are indeed equivalence preserving and having the creativity and experience to know which ones to apply and in what order.

What are the Equivalence-Preserving Operations on Equations?

If a = b, then f(a) = f(b) for any well-defined function f. If a and b are expressions containing a free-variable, then any value of that variable which satisfies a = b will also satisfyf(a) = f(b). In other words, if you find it useful, feel free to replace any equation with a new equation which is the result of applying any function to both sides of the original equation. Any solution to the original equation will also be a solution to the new equation.

If the function f is also one-to-one, then by definition, f(a) = f(b) \Rightarrow a = b so any solution of f(a) = f(b) will also be a solution to a = b. Thus applying f to both sides of an equation is equivalence-preserving. If f is not one-to-one, then in general, the operation is not equivalence-preserving.

In solving equation (1), we applied f(n) = n + 5, g(n) = n - 5x and h(n) = n/2 in that order. Since all three of the functions are one-to-one, we are assured that (1) and (4) are equivalent. If we had cause to apply a non-one-to-one function, then we should be vigilant for extraneous solution.

A More Interesting Example

Consider

(5) \sqrt{6x-2} - \sqrt{x+1} = 2

As I mentioned in the other post, these square roots are begging to be squared, but since there are two of them, one squaring will not be enough. Even though it’s not necessary to do so, it’s helpful to move one radical expression to the other side.

(6) \sqrt{6x-2} = 2 + \sqrt{x+1}

(7) 6x - 2 = 4 + 4\sqrt{x + 1} + x + 1 We squared!

(8) 5x - 7 = 4\sqrt{x+1}

(9) 25x^2 - 70x + 49 = 16x + 16 We squared again!

(10) 25x^2 - 86x + 33 = 0

(11) (25x - 11)(x - 3) = 0

So x \in \{\frac{11}{25}, 3\}

Since in the transition from (6)-(7) and again in the transition from (8)-(9) we had reason to apply the non-one-to-one function f(n) = n^2, we should be vigilant for extraneous solutions. [Note: since both sides of (6) are necessarily positive, applying f(n) = n^2 is equivalence-preserving, so no extraneous roots will be created there.] By checking back in the original equation, we see that 3 is a solution, but \frac{11}{25} is not. I am more or less content to leave it at that. But some may ask for more clarity as to exactly what happened and when, so let’s indulge them.

I will now list each equation in reverse order along with its solution set:

(11) (25x - 11)(x - 3) = 0                            \{\frac{11}{25}, 3\}

(10) 25x^2 - 86x + 33 = 0                             \{\frac{11}{25}, 3\}

(9) 25x^2 - 70x + 49 = 16x + 16                 \{\frac{11}{25}, 3\}

(8) 5x - 7 = 4\sqrt{x+1}                                     \{3\}

Since 5\cdot\frac{11}{25} - 7 = \frac{11}{5} - \frac{35}{5} = -\frac{24}{5} \neq 4\sqrt{\frac{11}{25} + 1} = 4\sqrt{\frac{11}{25} + \frac{25}{25}} = 4\sqrt{\frac{36}{25}} = 4\cdot\frac{6}{5} = \frac{24}{5}

So we have isolated the precise moment when the extraneous solution x = \frac{11}{25} is created and it appears exactly where we would expect it, in the transition from (8) to (9) as we replaced (8) with the result of applying the non-one-to-one function f(n) = n^2 to both sides.

More specifically, if x = \frac{11}{25}, (8) reads - \frac{24}{5} = \frac{24}{5}, which is false, but (9) reads (- \frac{24}{5})^2 = ( \frac{24}{5})^2, which is true. For this particular value of x, we squared both sides and replaced a false statement with a true statement. In retrospect, we can say that x =\frac{11}{25} is not a solution to (8) or to any previous equation in the solving sequence, but is a solution to (9) and thus to all subsequent equations in the solving sequence.

(7) 6x - 2 = 4 + 4\sqrt{x + 1} + x + 1                          \{3\}

Since both sides of (7) are positive when x = 3, it does not surprise us that,

(6) \sqrt{6x-2} = 2 + \sqrt{x+1}                               \{3\}

(5) \sqrt{6x-2} - \sqrt{x+1} = 2                                \{3\}

By fully analyzing the logic behind each step of our equation replacement sequence, we not only:

  • confirm that x = 3 is a solution and that x = \frac{11}{25} is not and
  • understand that squaring both sides may produce an extraneous solution

but also

  • isolate the precise step in the solving sequence in which this extraneous solution was created answering the why, how, and when for this problem
  • confirm that the non-solution status of x = \frac{11}{25} is not merely due to an error of algebra or arithmetic, but is a direct result of that fact that this value produces an equation (8) of the form a = -a

That last point is crucial in distinguishing the phenomenon of extraneous roots from the phenomenon of user error in algebra or arithmetic. If our equation solving sequence consists solely of equivalence-preserving operations, we do not even need to check to see if solutions to our terminal equation are also solutions to our original equation. If we do decide to check, perhaps out of an abundance of caution, and find a discrepancy, then user error must be to blame.

On the other hand, if a solver does employ solution-set-enlarging operations in the solving sequence and finds that a solution to the terminal equation is not a solution to the original equation, is this because the solution is extraneous or due to user error? One could perform an analysis like I did above and confirm that the non-solution is not due to user error, but instead to the logic of the process.

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.