function

f(x) = x^2 - 4x +1, f(a) = f(b) = 0
a) a² - 4a + 1
a² - 4a + 1 = f(a) = 0
b² - 4b + 1 = f(b) = 0

b) a, b are the roots of the equation f(x) = x² - 4x + 1 = 0
a + b = 4

c) f(a+b) = f(4) = 4² - 4*4 + 1
= 1.




2006-12-31 11:57:29 補充：
For part (b), we may evaluate by the steps below.a² - 4a 1 = b² - 4b 1 = 0a² - 4a = b² - 4ba² - b² = 4(a-b)(a b)(a-b) = 4(a-b)a b = 4, since a =/= b, a-b =/= 0

a

f(a)= a^2 - 4a +1=0

F(b)=b^2 - 4b +1=0

b)

as a, b are roots of f(x) (when we put a,b into f(x), f(x) touches x-axis)
so sum of roots is 1/1=1
so a+b=1


C)
f(1)=1^2-4*1+1=-2

2006-12-31 11:59:25 補充：
比你快一步/__\