How to Use the Discriminant

Solve for k, given Y_Parabola = 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, Y_Parabola = Y_Line. Therefore Y_Parabola – Y_Line = 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 Y_Parabola = 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.