Question

关于x的方程x^2+(4+i)x+4+mi (m属于R）有实数根n ,则n +mi=?

Answer 1

代入：

n^2+(4+i)n+4+mi

=n^2+4n+4+(n+m)i

=0

则n+m=0,n^2+4n+4=0

n=-2,m=2

n+mi=-2+2i