方程组y=m(x+2)-4,x^2+y^2=4有实 - 知识问答

方程组y=m(x+2)-4,x^2+y^2=4有实数解(x≥0),则m的取值范围是

1个回答

y=mx+(2m-4)
代入
(1+m²)x²+2m(2m-4)x+(2m-4)²=4
(1+m²)x²+4(m²-2m)x+4(m²-4m+3)=0
所以判别式△>=0
16(m^4-4m³+4m²)-16(m^4-4m³+3m²+m²-4m+3)>=0
4m-3>=0
m>=3/4

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