MATHS緊急,要STEPS

Consider a quadratic function f(x)=x^2-4x+1.Suppose f(a)=f(b)=0,where a unequal b.
a)Evaluate a^2 -4a +1 and b^2-4b+1.
f(a)=a^2-4a+1=0
f(b)=b^2-4b+1=0
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b)Hence find the value of a+b

a^2-4a+1=b^2-4b+1
a^2-4a=b^2-4b
a^2-b^2-4a+4b=0
(a+b)(a-b)-4(a-b)=0
(a+b-4)(a-b)=0
(a+b-4)=0 or (a-b)=0
(a+b)=4 or a=b(rejected, as a unequals b)
so the value of a+b=4
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c)Find f(a+b)
f(a+b)
=f(4)
=4^2-4(4)+1(base on f(x)= x^2-4x+1)
=16-16+1
=1

a)since f(x)=x^2-4x+1, so..
f(a)=a^2-4a+1=0
f(b)=b^2-4b+1=0

b)since (a), so..
a^2-4a+1=b^2-4b+1
a^2-4a =b^2-4b
a^2-b^2 -4a+4b=0
a^2-b^2 -4(a-b)=0
since x^2-y^2=(x-y)(x+y),so..
(a-b)(a+b) -4(a-b)=0
(a-b) [(a+b)-4]=0
(a-b) [(a+b)-4]/(a-b)=0/(a-b)
(a+b)-4=0
a+b=4

c)since f(x)=x^2-4x+1, so..
f(a+b)=(a+b)^2-4(a+b)+1
since a+b=4
then f(a+b)=f(4)
f(4)=4^2-4(4)+1
=16-16+1
=1