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.