x²+2x-11-k(x-3)=0x²-(k-2)x+3k-11=0方程有实根,判别式≥0[-(k-2)]²-4(3k-11)≥0k²-16k+48≥0(k-4)(k-12)≥0k≥12或k≤4设两根分别为x1,x2,由韦达定理得x1+x2=k-2x1x2=3k-11两根都>2x1+x2>2+2=4...