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

Answer 2

方程有实根,则△≥0
4(m+1)²-4(3m²+4mn+4n²+2)≥0
整理,得:2m²-2m+4mn+4n²+1≤0
(m²-2m+1)+(m²+4mn+4n²)≤0
(m-1)²+(m+2n)²≤0
m-1=0
m+2n=0
解得:
m=1,n=-0.5

Answer 3

2(m+1)×2(m+1)-4×(3m^2+4mn+4n^2+2)≥0
4mm+8m+4-12mm-16mn-16nn-8≥0
2m-1-2mm-4mn-4nn≥0
(m+2n)(m+2n)+(m-1)(m-1)≤0
m=1，n=-0.5

Answer 4

解：①当x≠0时，关于X的一元两次方程x²+2(m+1)x+3m²+4mn+4n²+2=0有实根
∴△=[2(m+1)]²－4(3m²+4mn+4n²+2)≥0
4m²+8m+4－(12 m²+16mn+16n²+8) ≥0
4m²+8m+4－12 m²－16mn－16n²－8≥0
2m²+4mn－2m+4n²+1≤0
(m+2n) ²+(m-1) ²≤0
∵(m+2n) ²≥0，，(m－1)²≥0，
∴ 只有当m+2n=0, m-1=0
∴m=1 ，n=-1/2