quadratic equation (form 4)

If the graph of y=(x-4p)(x-2)+p has only one x-intercept.
(a) Find the possible value(s) of p,
(b) find the x-intercept of the graph for each value of p in (a)


(a)
The graph: y = (x - 4p)(x - 2) + p ..... (1)
x - axis: y = 0 ..... (2)

Solve the above simultaneous equations to find the intersect of the graph and the a-axis.
(1) = (2):
(x - 4p)(x - 2) + p = 0
x(x - 2) - 4p(x - 2) + p = 0
x2 - 2x - 4px + 8p + p = 0
x2 - (2 + 4p)x + 9p = 0

Since there is only one x-intercept, determinant Δ = 0
[-(2 + 4p)]2 - 4(1)(9p) = 0
4 + 16p + 16p2 - 36p = 0
16p2 - 20p + 4 = 0
4p2 - 5p + 1 = 0
(p - 1)(4p - 1) = 0
p = 1 or p = 1/4


(b)
When p = 1:
x2 - [2 + 4(1)]x + 9(1) = 0
x2 - 6x + 9 = 0
(x - 3)2 = 0
x = 3 (double root)

When p = 1/4:
x2 - [2 + 4(1/4)]x + 9(1/4) = 0
x2 - 3x + 9/4 = 0
x2 - 2x(3/2) + (3/2)2 = 0
(x - 3/2)2 = 0
x = 3/2 (double root)

x-intercept of the graph: (3, 0) or (3/2, 0)