一條maths mc 值20分

a + b = 2 [sum of roots]
ab = -6 [product of roots]

(a-b)^2 = a^2 - 2ab + b^2
=(a^2-2ab+b^2)+4ab
=(a-b)^2 + 4ab
=2^2 + 4*-6
=-20

2007-04-21 18:49:36 補充：
some mistake in the middleThe corrected one:(a-b)^2 = a^2 - 2ab + b^2=(a^2+2ab+b^2)+4ab=(a+b)^2 + 4ab=2^2 + 4*-6=-20However, the answer is the same

"完美板"

先講一講點搵二次方程o既 sum of roots 同 product of roots
let p*x^2+q*x+r=0
sum of roots = -q/p
product of roots = r/p

所以, 題目 x^2-2*x+6=0
sum of roots = -(-2)/1=2
product of roots = 6/1=6 (呢度就係上面某位唔小心o既地方)

(a-b)^2 = a^2 - 2ab + b^2
=(a^2+2ab+b^2)-4ab
=(a+b)^2 - 4ab
=2^2 - 4*6
=-20

同埋上面有一個人完全copy人地錯o既答案, 我而家去檢舉佢"

2007-05-03 00:34:00 補充：
有份投票都冇腦, 做錯仲投佢一票

首先 ( a - b ) ( a - b )
= a^2 - 2ab + b^2
= a^2 + b^2 - 2ab
= a^2 + 2ab + b^2 - 2ab - 2ab ...... + 一個 2ab 比 佢...等佢可以 變成 ( a+b ) ( a+b )
=( a + b ) ( a + b ) - 4ab ........再用返 兩根之和 a+b = -(-2)/1 同 兩根之積 ab = 6/1
= (2) (2) - 4(6)
= 4-24
= -20

hope that can help you laaa

a + b = 2 [sum of roots]
ab = -6 [product of roots]

(a-b)^2 = a^2 - 2ab + b^2
=(a^2-2ab+b^2)+4ab
=(a-b)^2 + 4ab
=2^2 + 4*-6
=-20

the equation is not work .

唔明你條問題...用中文好唔好??

i dont undersatand ur question..."
a , b are the roots of the equation x 二 次 mean wt?