来自董彬的问题
若方程x2+2(m+1)x+3m2+4mn+4n2+2=0有实数根,则实数m=(),实数n=()
若方程x2+2(m+1)x+3m2+4mn+4n2+2=0有实数根,则实数m=(),实数n=()
1回答
2020-01-21 00:39
若方程x2+2(m+1)x+3m2+4mn+4n2+2=0有实数根,则实数m=(),实数n=()
若方程x2+2(m+1)x+3m2+4mn+4n2+2=0有实数根,则实数m=(),实数n=()
若方程x2+2(m+1)x+3m2+4mn+4n2+2=0有实数根,判定式Δ≥0.
Δ=[2(m+1)]^2-4(3m^2+4mn+4n^2+2)≥0
4(m+1)^2-4(3m^2+4mn+4n^2+2)≥0
(m+1)^2-(3m^2+4mn+4n^2+2)≥0
m^2+2m+1-3m^2-4mn-4n^2-2≥0
-m^2+2m-1-m^2-4mn-4n^2≥0
-(m^2-2m+1)-(m^2+4mn+4n^2)≥0
-(m-1)^2-(m+2n)^2≥0
(m-1)^2+(m+2n)^2≤0
左边的两个平方数都只能是非负数,所以都只能是0:
m-1=0,m+2n=0
m=1
n=-1/2