已知x1和x2是方程(k²-1)x²-6(3k-1)x+72=0的两正根,且(x1-1)(x2-1)=4,求k的值
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已知x1和x2是方程(k²-1)x²-6(3k-1)x+72=0的两正根,且(x1-1)(x2-1)=4,求k的值
已知x1和x2是方程(k²-1)x²-6(3k-1)x+72=0的两正根,且(x1-1)(x2-1)=4,求k的值
已知x1和x2是方程(k²-1)x²-6(3k-1)x+72=0的两正根,且(x1-1)(x2-1)=4,求k的值
显然,x1+x2=6(3k-1)/(k²-1),x1x2=72/(k²-1)
(x1-1)(x2-1)=x1x2-(x1+x2)+1=(78-18k)/(k²-1)+1=4
k²+6k-27=0
k=3或k=-9
由于x1,x2均为正数,故x1+x2,x1x2均为正
代入验证得k=3