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., y_L = y_P. 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., x_L = x_P. 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 x_L = x_P is likely to be easier than working with y_L = y_P.

Averages

What does the word average signify?  Did you know Wikipedia lists at least 14 different kinds of averages? (http://en.wikipedia.org/wiki/Average)

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. 

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!