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標題: 二次方程問題. [打印本頁]

作者: 富商.豪 時間: 09-9-3 06:35 PM 標題: 二次方程問題.

eqt:(14m-1)x^2 -2mx + 1 =0

given that a and b are the roots of this eqt.

if 2a-b=3ab,
find m.

作者: 98102257 時間: 09-9-4 10:15 AM

a+b=2m/(14m-1) ... (1)
ab=1/(14m-1) ... (2)

2a-b=3ab
2a-b=3/(14m-1) ... (3)

(1)+(3)
3a=(2m+3)/(14m-1)
a=(2m+3)/3(14m-1) 代入 (1)
b=(4m-3)/3(14m-1)

所以 ab=(2m+3)(4m-3)/9(14m-1)^2=1/(14m-1) by (2)
(2m+3)(4m-3)=9(14m-1)
8m^2+6m-9=126m-9
m^2-15m=0
m=0 or 15

如果 m=0
a=-1, b=1

如果 m=15
a=1/19, b=1/11

作者: 富商.豪 時間: 09-9-4 06:22 PM

哦-0-明白了,,
唔該哂~

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