Here are two problems from a diagnostic quiz given to a Calculus 1 for engineers class this semester at University of Maryland. Thanks to Leigh Lieberman for sending me these!

1. 1.000002 × 1.000003

2. (√3 − √2)^4 x (√3 + √2)^4

Most of the students were unable to calculate the correct answers. Can you?

**Challenge problems:**

3. The most typical answer given by these students to the second problem was 65. Explain what they did wrong.

4. Relate it to lack of emphasis on the Distributive Law and the Special Rules of polynomial multiplication discussed in this week's Problems of the Week.

5. Use your answers to 3 and 4 to predict what is likely to have been the most typical answer to problem 1.

## Saturday, February 16, 2013

### The Distributive Law and the Special Rules Revisited: what instant polynomial pattern recognition can do for you

Labels:
math,
standard algorithms

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## 9 comments:

MATHCOUNTS uses problems like that in the Countdown round quite often. When a student gets the answer in a couple seconds, everyone gasps because they don't realize how easy they are if you know the patterns ...

Wow, it took me awhile to figure out how you would get any number other than 1 for the second one, but I think I have it figured out. They're distributing the ^4, aren't they?

(√3 − √2)^4 x (√3 + √2)^4

((√3)^4 − (√2)^4) x ((√3)^4 + (√2)^4)

(9 - 4) x (9 + 4)

5 * 13

65

The most amusing thing to me is that when I plugged that into my calculator application, it came up with 8. After adding gratuitous parentheses with no change in result, I finally figured out that the calculator was interpreting ^ as something other than "raise to the power of", which probably means it was doing an XOR or something.

The first problem is pretty easy, but the second took me a while--mostly because I was doing one stupid thing. Still, isn't the answer to #2:

0.000104145...

Or, more accurately:

4801-1960*√6

Step 1, do the square of the parentheses:

(√3 − √2)^2 = 3 + 2 - 2√3√2

= 5 - 2√6

Now, do it again to get ^4:

(√3 − √2)^4 = (5 - 2√6)^2 = 25 + 24 - 20√6

= 49 -√6

Now do it one more time to get ^8:

(√3 − √2)^8 = (49 -√6)^2 = 49^2 + 2400 - 49*40*√6

= 4801 - 1960√6

I suspect back when math & science students were taught to use slide rules these would have been considered trivial jokes, a secret handshake among those that "wear the holster" and those that don't.

Nowadays they really are a test of how well one knows their algebra.

Were they allowed calculators for problem 1? Were they expected to provide the exact mathematical answer or to round to the correct number of significant figures? or was either answer acceptable.

The answer to the second question is "x" which most students would get wrong because they wrongly think that "x" is a multiplication symbol.

You get 65 on the second problem if you incorrectly apply the 4th power to each square root, getting 3 squared and 2 squared, so (9-4)(9+4) or 5*13 = 65.

The first problem is interesting because a calculator probably won't show enough significant digits to do the problem. Although if you try it with fewer zeroes and follow the pattern, you can do it.

I suspect the most common error on the first one is something like 1.000006

I suspect the most common error on the first one is something like 1.000006EEEEEEEEE

Hehe, those that know will know.

Problem 1 is easy to do on paper if you break it down like:

(1 + 2/1,000,000) * (1 + 3/1,000,000)

= 1 + 5/1,000,000 + 6/1,000,000,000,000

= 1.000005000006

Yes, I didn't see the plus sign in the second parentheses. That'll larn me to look more carefully. Yes, the answer is 1.

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