## Monday, June 20, 2011

### How to Use the Discriminant

Solve for k, given YParabola = 1x2 + (2k + 10)x + k2 + 5,  tangent to the x-axis.
Mary says “Obviously”, b2 – 4ac = 0              (Johnny says, "This may not be obvious.  I will provide a note below.)

 (2k + 10)2 – 4(1)( k2 + 5) = 0 Substitution of  a = 1, b = 2k + 10 and c = k2 + 5 4k2 + 40k + 100 –  4k2  – 20 = 0 Expand 40k = -80 k = -2 Done.

Note:  At the point where a line is tangent to a parabola, YParabola = YLine. Therefore YParabola YLine = 0.  But in this case the tangent line is the x-axis (given) so the line is simply y = 0.  (or, y = 0x + 0 if you want to be precise!) Since the gradient of y = 0 must be m = 0, we know the tangent is horizontal. This only happens at a stationary point.  But a parabola only has one stationary point (the maximum or minimum).  In this case the stationary point touches the x-axis so it is also the only root of the parabola.  Consequently the discriminant, b2 – 4ac is 0.  (Recall that, if there are two real roots b2 – 4ac > 0, if there are no real roots b2 – 4ac < 0.)
Calculus has only been used in the background to understand the problem.   We never found the first derivative.  This suggests an alternative but more complicated approach.
Alternate method:
m = dy/dx = 2x + 2k + 10
As noted above, m = 0  (The tangent is horizontal).
0 = 2x + 2k = 10
0 = 2(x + k + 5)
0/2 = 0 = x + k + 5
x =  – k – 5
Substitute this x into YParabola = 0             (The tangent is the x-axis which is the line y = 0).
1x2 + (2k + 10)x + k2 + 5 = 0
(–k – 5)2 + (2k + 10) (–k – 5) + k2 + 5 = 0
k2 + 10k + 25 – 2k2 – 10k –10k – 50 + k2 + 5 = 0
20 = -10k
k = -2          As before. Undoubtedly there exist even more complicated solutions but Johnny will stop here.