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標題:
Form 3 maths pls help
發問:
a+b=4, ab=3, find the value of (a-b)^2. the answer should be 4. thanks.
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
其他解答:
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
Form 3 maths pls help
發問:
a+b=4, ab=3, find the value of (a-b)^2. the answer should be 4. thanks.
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最佳解答: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
其他解答:
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
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