Re: [中學] 三次多項式三根成等差
※ 引述《magiclass (課堂上玩數學)》之銘言:
: 三次多項式x^3+ax^2+bx+2=0的三根成等差,
: 且a,b皆為整數。
: 求a=? b=?
設三根為r-d,r,r+d
由根與係數關係可得
3r=-a...........(1)
3r^2-d^2=b......(2)
r(r^2-d^2)=-2...(3)
接著將r,d消掉,可得a,b關係式
由(3)->d^2=r^2+2/r
代回(2)->b=2r^2-2/r...(4)
由(1)->r=-a/3
代回(4)->b=6/a+2a^2/9
整理後得2a^3-9ab=-54
->a(2a^2-9b)=-54
a可能等於±1,±2,±3,±6,±9,±18,±27,±54
代回求b
整數解有(a,b)=(3,4),(6,9),(-3,0),(-6,7)
感謝推文提醒
d^2=r^2+2/r=a^2/9-6/a>=0
代回檢查(3,4)不合
答案剩下(6,9),(-3,0),(-6,7)
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※ 編輯: alice90426 (114.44.236.26), 12/05/2014 00:05:37
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