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 consists of 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 when 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 turned a false statement into 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.

Advertisements

One thought on “Extraneous Solutions – Part 2 of 3

  1. Will,

    I was giving you wholehearted “amens” on your first post, but I can’t give you as strong of an endorsement on this post. I still love you, though :-).

    If someone proves that \tan{x}-\sin{x}\cos{x}=\frac{\sin^3{x}}{\cos{x}} by writing the following work,

    \sin^2{x}=\sin^2{x}
    1-\cos^2{x}=\sin^2{x}
    \tan{x}-\tan{x}\cos^2{x}=\tan{x}\sin^2{x}
    \tan{x}-\sin{x}\cos{x}=\frac{\sin^3{x}}{\cos{x}},

    then I think we would agree that this constitutes a proof. The truth of each statement follows from the previous statement.

    I know you want to draw a line in the sand between this kind of algebraic work (a “proof”) and the kind of algebraic work you speak about in your post which doesn’t aim to prove anything, but simply to faithfully generate solutions. But I think these two seemingly disparate activities can be brought under the same roof, and even considered to be two tasks that are members of the same class of problems.

    I hear what you’re saying about how we actually “do” algebra, and how we don’t usually think of it in these philosophical terms. I agree. And in our classrooms, I’m sure you and I provide all the same sets of tools to our students for navigating all the tricky situations involving extraneous solutions. At the level of pedagogy, I think we agree. So the conversation I want to have is at the level of Philosophy of Math or Proof Theory or something like that, not pedagogy.

    Looking forward to #3 in your series,
    John

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s