Form 3 maths pls help

a + b = 4 ----- equation 1
ab = 3 ----- equation 2
From equation 2, a = 3 / b ----- equation 3
3 / b + b = 4 ----- substitute equation 3 into equation 1
3 + b^2 = 4b
b^2 - 4b + 3 = 0
(b - 3) (b - 1) = 0
b = 1 or b = 3

From equation 1, when b = 1, a = 4 - 1 = 3,
(a-b)^2= (3-1)^2 = 2^2 = 4

when b = 3, a = 4 - 3 = 1
(a-b)^2= (1-3)^2 = (-2)^2 = 4

2006-11-21 14:29:37 補充：
還有另一個計算方法︰(a-b)^2 = a^2 - 2ab b^2= a^2 2ab b^2 - 4ab= (a b)^2 - 4ab= 4^2 - 4*3= 16 - 12= 4

2006-11-21 14:31:32 補充：
還有另一個計算方法︰(a-b)^2 = a^2 - 2ab + b^2= a^2 + 2ab + b^2 - 4ab= (a+b)^2 - 4ab= 4^2 - 4*3= 16 - 12= 4

a+b=4, ab=3

(a-b)^2 = a^2 - 2ab + b^2
= a^2 + 2ab + b^2 - 2ab - 2ab
= (a+b)^2 - 4ab
= 4^2 - 4*3 (because a+b = 4, ab=3)
= 16 - 12
= 4