Question

一道整式题求解.已知关于x的一元二次方程 x^2+2(m+1)x+(3m^2+4mn+4n^2+2)=0有实根,则m,n

Answer 1

有实根

所以判别式大于等于0

[2(m+1)]^2-4(3m^2+4mn+4n^2+2)≥0

4m^2+8m+4-12m^2-16mn-16n^2-8≥0

-2m^2-4mn-4n^2+2m-1≥0

2m^2+4mn-2m+4n^2+1≤0

(m+2n)^2+(m-1)^2≤0

所以(m+2n)^2=0,(m-1)^2=0

m=1

n=-m/2=-1/2