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Saturday, February 18, 2012

Quadratics, a Discriminant Problem

The line y + 2x = p has no intersection with the parabola y2 = x + p. Prove that p < -1/24.
Whenever there is an intersection, the two y co-ordinates must be equal, i.e., yL = yP. According to the usual method, Mary lets y = p – 2x, so y2 = (p – 2x)2 = p2 – 4px +4x2.
y2 = y2, so p2 – 4px +4x2 = x + p
Then, 4x2 – 4px + p2 – x – p = 0
4x2 – (4p + 1) x + (p2 – p) = 0
Finally, Mary uses the discriminant, b2 – 4ac, with a =4, b = – (4p + 1) and c = p2   p.
b2 – 4ac < 0
(– (4p + 1))2 – 4(4)(p2   p) < 0
16p2 + 8p + 1 – 16p2 + 16p <0
24p < -1
p < -1/24

However, it must also be true that, whenever there is an intersection, the two x co-ordinates must be equal, i.e., xL = xP. Therefore Johnny writes:

Johnny also has a quadratic equation, and he also uses the discriminant. However, he has somewhat ‘nicer’ values for a, b and c. a = 1, b =½ and c = -1.5p.
b2 – 4ac < 0
(½)2 – 4(1)(-1.5p) < 0
¼ + 6p <0
6p < -¼
p < -1/24
If you believe that every second counts on the exam, it pays to notice that y2 = x + p is not our typical parabola, as it opens rightwards (instead of upwards or downwards).

This observation ought to give rise to the notion that working with xL = xP is likely to be easier than working with yL = yP.


What does the word average signify?  Did you know Wikipedia lists at least 14 different kinds of averages? (
The most common average is the arithmetic mean, but it is not always appropriate.  Basil runs 800 metres at 8 metres per second and another 800 m at 7 metres per second.  What is his average speed?  Many ‘good’ students blithely add 8 and 7 and then divide by 2 to get 7.5. wrong!
Let’s take an extreme case.  Extreme cases are an essential part of your mathematical toolbox. Suppose Basil runs his first 800 m at 8 m/s and his second 800 m at the glacially slow speed of 0. 0000…1 m/s.  (Imagine as many zeros as you like before the one). Is the average speed (8 + 0)/2 = 4 metres per second?  Certainly not!
Imagine another case.  I am driving my car at 100 km per hour for 0.99999 …9 hours (or 59.999…9 minutes) and for a brief fleeting instant at 200 km per h.  Is my average speed (100+200)/2 = 150 km/h?  Certainly not!
Instead of the ordinary mean (a + b)/2, what the student needs is a weighted mean (w1 × a + w2 × b)/( w1 + w2).   But, what are the appropriate weights?  The obvious (and correct answer) are the times spent at each speed.  Then we get (t1 × s1 + t2 × s2)/( t1 + t2) = (d1 + d2)/( t1 + t2) = total distance/total time, as taught by the physics teachers. 

Saturday, February 11, 2012

Unusual products

We all know that 9 × 7 is equal to 7 × 9. However, the following pairs of products appear unusual.

Example 1:

Example 2:

Usually, ‘reversed products’ are not the same:

Example 3:

What distinguishes the identical products from the non-identical products?  Do the factors offer any clues?
Here is another look at Example 1:
Note that 2 × 3 = 6 × 1, or in other words the product of the Units is the same as the product of the Tens.  When we do long multiplication our first ‘sub-product ‘(or should that be sub-total) is  U × U, followed by   U × T, then T × U and finally T × T. Therefore, when we ‘reflect’ 13 and 62 all of the ‘sub-products’ stay the same, although the middle pair U × T and T × U are in a different order.

For Secondary Students:
We can use algebra to give the general pattern.  Let A, B, C and D represent any of the digits from 0 to 9.  Please note that AB  and CD are not products but concatenations (or strings). That is, in polynomial notation, AB would be written as 10 A + 1 B and likewise CD would be 10 C + 1 D.

Extending the Pattern

Now that we have some understanding of the pattern, it is natural to ask if it can be extended to three digit numbers.    Have a go!