原方程可化为
(k-4)(k-2)x2+(2k2-6k-4)x+(k-2)(k+2)=0,
〔(k-4)x+(k-2)〕〔(k-2)x+(k+2)〕=0,
∵(k-4)(k-2)≠0,∴x1=-(k-2)/(k-4)=-1-2/(k-4),
x2=-(k+2)/(k-2)=-1-4/(k-2).
∴k-4=-2/(x1+1),k-2=-2/(x2+1).(x1≠-1,x2≠-1)消去k,得
x1x2+3x1+2=0,∴x1(x2+3)=-2.由于x1、x2都是整数,
∴x1=-2,x2+3=1; x1=1,x2+3=-2; x1=2,x2+3=-1.
∴x1=-2,x2=-2; x1=1,x2=-5; x1=2,x2=-4.∴k=6,3,10/3.经检验,k=6,3,10/3满足题意
设关于x的二次方程(k^2-6k+8)x^2+(2k^2-6k-4)+k^2=4的两个根都是整数,求k的取值
设关于x的二次方程(k^2-6k+8)x^2+(2k^2-6k-4)+k^2=4的两个根都是整数,求k的取值
数学人气:593 ℃时间:2020-02-01 10:33:57
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