x²+2(a+1)x+3a²+4ab+4b²+2=0. 有实数根、、求a和b的范围、、(请写出过程)
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/02 16:54:49
![x²+2(a+1)x+3a²+4ab+4b²+2=0. 有实数根、、求a和b的范围、、(请写出过程)](/uploads/image/z/2548437-69-7.jpg?t=x%26%23178%3B%2B2%EF%BC%88a%2B1%29x%2B3a%26%23178%3B%2B4ab%2B4b%26%23178%3B%2B2%3D0.+%E6%9C%89%E5%AE%9E%E6%95%B0%E6%A0%B9%E3%80%81%E3%80%81%E6%B1%82a%E5%92%8Cb%E7%9A%84%E8%8C%83%E5%9B%B4%E3%80%81%E3%80%81%EF%BC%88%E8%AF%B7%E5%86%99%E5%87%BA%E8%BF%87%E7%A8%8B%EF%BC%89)
x²+2(a+1)x+3a²+4ab+4b²+2=0. 有实数根、、求a和b的范围、、(请写出过程)
x²+2(a+1)x+3a²+4ab+4b²+2=0. 有实数根、、求a和b的范围、、(请写出过程)
x²+2(a+1)x+3a²+4ab+4b²+2=0. 有实数根、、求a和b的范围、、(请写出过程)
方程有实数根
所以判别式≥0
4(a+1)²-4(3a²+4ab+4b²+2)≥0
a²+2a+1-(3a²+4ab+4b²+2)≥0
-2a²+2a-4ab-4b²-1≥0
(-a²+2a-1)-(a²+4ab+4b²)≥0
-(a-1)²-(a+2b)²≥0
(a-1)²+(a+2b)²≤0
所以a-1=0
a+2b=0
所以a=1,b=-1/2
有实根
4(a+1)^2-4(3a^2+4ab+4b^2+2)≥0
a^2+2a+1-3a^2-4ab-4b^2-2≥0
-2a^2+2a-1-4ab-4b^2≥0
2a^2-2a+1+4ab+4b^2≤0
(a^2-2a+1)+(a^2+4ab+4b^2)≤0
(a-1)^2+(a+2b)^2≤0
平方和小于0不成立
所以只能等于0
所以(a-1)^2=0,(a+2b)^2=0
a-1=0,a+2b=0
a=1,b=-1/2