Question

已知x的一元二次方程x²+2(m+1)+3m²+4mn+4n²+2=0有实数根,则3m

Answer 1

4(m+1)^2-4(3m^2+4mn+4n^2+2)≥0

m^2+2m+1-3m^2-4mn-4n^2-2)≥0

-2m^2-4mn-4n^2+2m-1≥0

2m^2+4mn+4n^2-2m+1≤0

m^2+4mn+4n^2+m^-2m+1≤0

(m+2n)^2+(m-1)^2≤0

平方和非负,所以

(m+2n)^2=0,（m-1)^2=0

m=-2n,m=1

m=1,n=11/2

3m²+2n²=3*1^2+2*(-1/2)^2=7/2

3m²+2n²的值为 7/2