3y^2+6y+1=3(y+m)^2+n
what is the value of n?
F.3 equation
2012-10-28 8:41 am
回答 (2)
2012-10-28 9:11 am
✔ 最佳答案
3y² + 6y + 1 = 3(y + m)² + n3y² + 6y + 1 = 3(y² + 2my + m²) + n
3y² + 6y + 1 = 3y² + 6my + (3m² + n)
Compare the y terms on the both sides :
6y = 6my
m = 1
Compare the constant terms on the both sides :
3m² + n = 1 ...... [1]
Put m = 1 into [1] :
3(1)² + n = 1
n = -2
參考: wanszeto
2012-10-28 11:43 pm
n=(4ac-b^2)/4a
n=(4*3*1-6*6)/(4*3)
n=-2
n=(4*3*1-6*6)/(4*3)
n=-2
收錄日期: 2021-04-20 13:19:14
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https://hk.answers.yahoo.com/question/index?qid=20121028000051KK00027