Question

若关于x的一元二次方程x^2+4mx+5m^2-2mn+n^2=0有一个根为4,则m+n=

Answer 1

x^2+4mx+5m^2-2mn+n^2=0有一个根为4

16+16m+5m^2-2mn+n^2=0

4(4+4m+m^2)+m^2-2mn+n^2=0

(m-n)^2+4(m+2)^2=0

∵(m-n)^2≥0

4(m+2)^2≥0

∴m=n

m=-2

∴m+n=-4