MATHS

f [( √2006 + 1) / 2] = 4 [( √2006 + 1) / 2] ^2 - 4 [( √2006 + 1) / 2] - 2006
= (√2006 + 1)^2 - 2(√2006 + 1) - 2006
= 2006 + 2√2006 + 1 - 2√2006 - 2 - 2006
= -1
or

f(x) = 4x^2 - 4x - 2006 = 4x^2 - 4x + 1 - 2006 - 1
= (2x - 1)^2 - 2006 - 1
= (2x - 1 - √2006)(2x - 1 + √2006) - 1

=> f[ (√2006 + 1)/2 ] = [2(√2006+1)/2 - 1 - √2006][2(√2006+1)/2 - 1 + √2006] - 1
= 0 * [2(√2006+1)/2 - 1 + √2006] - 1
= -1

2006-11-14 19:38:20 補充：
for the 2nd part, we use the formula x^2 - y^2 = (x y)(x-y)

Let a = (1 + sqrt[2006]) / 2
b = (1 - sqrt[2006]) / 2

a + b = 1
ab = 1/4 (1 - 2006)

The quadratic equation with roots a and b is
x^2 - x + 1/4 (1 - 2006) = 0
4x^2 - 4x - 2006 + 1 = 0
f(x) + 1 = 0
Since a is a root of f(x) + 1 = 0
f(a) + 1 = 0
f(a) = -1

2006-11-14 19:35:14 補充：
the answer given from the person above me is a little bit slow.

2006-11-14 19:36:24 補充：
This question is twisted.

f [( √2006 + 1) / 2]
=4[( √2006 + 1) / 2]²-4[( √2006 + 1) / 2]-2006
=4[(2006+2√2006+1)/4]- 2(√2006 + 1) -2006
=2007+2√2006-2√2006 -2 -2006
=2007-2-2006
= -1