方程有兩根,二次項(xiàng)系數(shù)≠0,方程判別式△≥0,兩根均為負(fù),兩根之和0.
k-2≠0 k≠2
△≥0 [-(3k+6)]²-4(k-2)(6k)≥0
5k²-28k-12≤0
(k-6)(5k+2)≤0
-2/5≤k≤6
設(shè)兩根分別為x1,x2,由韋達(dá)定理得
x1+x2=(3k+6)/(k-2)
x1x2=6k/(k-2)
(3k+6)/(k-2)