## 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.