若方程x^2＋2(m+1)x+3m^2+4mn+4n^2+2=0有实根,则实数m,n等于多少?

Question

若方程x^2＋2(m+1)x+3m^2+4mn+4n^2+2=0有实根,则实数m,n等于多少?

Answer 1

因为关于X的一元两次方程x^2+2(m+1)x+(3m^2+4mn+4n^2+2)=0有实根
所以△=〔2(m+1)〕^2-4(3m^2+4mn+4n^2+2)≥0
4m^2+8m+4-(12 m^2+16mn+16n^2+8) ≥0
4m^2+8m+4-12 m^2-16mn-16n^2-8≥0
合并同类项,整理得
2m^2+4mn-2m+4n^2+1≤0
(m+2n) ^2+(m-1) ^2≤0
m=1
n=-1/2