m取何整数时方程(m^2-1)x^2-6(3m-1)x+72=0的根是正整数

问题描述:

m取何整数时方程(m^2-1)x^2-6(3m-1)x+72=0的根是正整数

整理分解因式 得:
(mx-x-6)(mx+x-12) = 0
所以 mx-x-6 = 0 或 mx+x-12=0
mx-x-6 = 0 ==> (m-1)x = 6 => m-1 必须是6的正因子 => m = 2,3,4,7
mx+x-12 = 0 ==> (m+1)x = 12 => m+1 必须是12的正因子 => m = 0,1,2,3,5,11
综合上面,m = 0,1,2,3,4,5,7,11