.若方程：x^2+2(m+1)x+3m^2+4mn+4n^2+2=0有实数根,则实数m=?；且实数n=?
问题描述：
.若方程：x^2+2(m+1)x+3m^2+4mn+4n^2+2=0有实数根,则实数m=?；且实数n=?
答
有实数根
△>=0
4(m+1)²-4(3m²+4mn+4n²+2)>=0
m²+2m+1-3m²-4mn-4n²-2>=0
2m²+4mn+4n²-2m+1所以(m²-4mn+4n²)+(m²-2m+1)(m-2n)²+(m-1)²所以只有m-2n=0,m-1=0
所以
m=1
n=m/2=1/2
答
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-2...