设公共解是a
则a²-(k+2)a+12=0 (1)
2a²-(3k+1)a+30=0 (2)
(1)×2-(2)
-2(k+2)a+(3k+1)a+24-30=0
(k-3)a=6
a=6/(k-3)
代入a²-(k+2)a+12=0
36/(k-3)²-6(k+2)/(k-3)+12=0
两边乘(k-3)²)/6
6-(k-3)(k+2)+2(k-3)²=0
k²-11k+24=0
k=3,k=6
a=6/(k-3),所以k≠3
k=6
公共解a=2