当a,b为何值时,方程x²+2(1+a)x+(3a²+4ab+4b²+2)=0有实数根
人气：412 ℃ 时间：2019-10-19 12:40:45
解答
∵方程有实数根∴判别式大于等于0△=[2(1+a)]²-4(3a²+4ab+4b²+2)=[4(1+2a+a²)]-4(3a²+4ab+4b²+2)=(4+8a+4a²)-(12a²+16ab+16b²+8)=4+8a+4a²-12a²-16ab-16b&#...
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