Saturday, May 2, 2009

A Tale of Two Equations: Balancing Procedures and Conceptual Understanding


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The following is intended for all students in 2nd year algebra. Your stronger students should not find these overly challenging but there is more here than meets the eye. The purpose here is to demonstrate how we can review procedures AND develop deeper understanding of important mathematical ideas in the same lesson. The graphing calculator can be used to enhance the lesson by employing multiple representations (Rule of Four) to reinforce the essential ideas.

Note: Finding the solutions is only the tip of the iceberg. Understanding WHY one equation must have finitely many solutions and the other must have infinitely many solutions is the bigger idea here...


Equation 1:
(x-1)(x-2) = (1-x)(2-x)

Equation 2:
(x-1)(x-2)(x-3) = (1-x)(2-x)(3-x)

Click on Read More... for solutions and further discussion.

Equation 1: All real numbers
Equation 2: {1,2,3}


Would most of your students eliminate parentheses in the first equation and solve by traditional methods? Even though the left and right sides of the equation appear similar, it is reasonable to expect they will distribute and solve since that is what they're used to doing. This is fine and the standard procedure should be reviewed.

Assuming students will not make "careless" mechanical errors in distributing, they should obtain:
x2 - 3x + 2 = 2 - 3x + x2.
This generates a nice discussion of an "identical equation" or identity since the left and right sides are mathematically equivalent (if they recognize that!). The instructor may or may not want to continue the mechanical approach of moving all terms to one side producing 0 = 0 to reinforce that the equation is satisfied by all real numbers. Your stronger student will not have much difficulty with this.

Before moving on to the 2nd equation, we can develop a
deeper conceptual understanding by asking students to approach the problem another way. We know that some students will wonder about the form of the original equation. Could we have predicted that the two sides would be identical without removing parentheses? Could we also have determined by inspection that both x = 1 and x = 2 are solutions? Asking them to revisit the original equation to see this is critical. Now what about trying some other real number, say x = 5. This should strongly suggest that all real numbers will satisfy the equation. Using the graphing calculator will also drive this point home visually. Store the left side of the equation Y1 and the right side of the equation in Y2. Change the appearance of Y2 (make it bold for example) and have them observe on the viewscreen that the graphs are identical.

So, how come the 2nd equation only has 3 solutions! I'll leave that to my readers to elaborate on...
How can we generalize this?

These kinds of lessons seem to involve way too much overhead, stealing so much valuable time away from other content. BUT these are precisely the kinds of problems students are expected to grapple with in Japan and other countries. Do you really believe "Less is More?"


Maria Miller said...

Here's a way to solve the second:

First note the obvious solutions x = 1, 2, 3.

Then, since x - 1 = (-1)(1-x), we can rewrite it as

(x-1)(x-2)(x-3) = (-1)(x-1)(-1)(x-2)(-1)(x-3)


(x-1)(x-2)(x-3) = (-1)(x-1)(x-2)(x-3)

Now would be time to cancel... which means DIVIDING by (x - 1), which you can do assuming x is not 1.

So assuming x is not 1, 2, or 3, we divide by (x-1)(x-2)(x-3) and get

1 = -1

So there won't be other solutions besides the x = 1, 2, 3 already found.

Maria Miller said...

Oh, another note on the first equation. Using the same "trick" of writing 1 - x as (-1)(x - 1) we get:

(x-1)(x-2) = (-1)(x-1)(-1)(x-2)

Now we can cancel (assuming x is not 1 or 2)... and since the two -1's on the right give 1, we get a true equation

1 = 1.

Maria Miller said...

Sorry for these multiple comments.

Then once you notice that you either are left with a -1 or 1 on the right side, you can generalize as to what happens if you have an even number or an odd number of factors in this equation "game".

Dave Marain said...

Wonderful comments, Maria.
It's nice to know that some out there appreciate WHY I am writing these questions. I have always believed that routine problems reinforce skill and lower-level reasoning but only non-routine problems will develop higher-order thinking. From textbooks I've seen from other nations (Russian, Korean, Singaporean, etc.), I see both routine and more challenging exercises in the regular homework sections. Furthermore, the model problems contain some more difficult problems. We seem to be reluctant in this country to include challenges in our textbooks. They tend to be in ancillary materials or an "Extra for Experts" section. This has to change.

I gave one of these questions to my small group. Some got it right away by a couple of different methods (including expanding!) but then I asked them to consider the equation A = -A. Anything this "simple" always involves conceptual thinking and a couple of them just stared at me. I asked the question "verbally" (multiple representations): "What is the only number which equals to its opposite?" Oh, of course...

Of course, one can mechanically 'solve' A = -A by avoiding "cancelling", just by adding 'A' to both sides:
2A = 0.
This same argument can be applied to your method:
(x-1)(x-2)(x-3) = -(x-1)(x-2)(x-3) can be transformed to
2(x-1)(x-2)(x-3) = 0 to avoid a discussion of 'cancelling' variables. However, I think a discussion of the ideas behind cancelling is very valuable. You elucidated this beautifully.

Back to the original (x-1)(x-2) = (x-1)(x-2). Even without 'cancelling', I like the idea of having students understand the meaning of an "identical equation" or identity. I feel they should recognize that the equation A = A leads to all real numbers as the solution, without necessarily going to 1 = 1.