方程(4+3i)x^2+mx+4-3i=0有实数根,求复数m的模的最小值

(4+3i)x^2+mx+4-3i=0有实数根,
显然x=0不符合题意
x不等于0时,
m=-(4+3i)x-(4-3i)/x
=(-4x-4/x)+(-3x+3/x)i
则|m|=根号下[(-4x-4/x)^2+(-3x+3/x)^2]
=根号下(25x^2+25/x^2+14)
>=根号下[2根号下(25x^2*25/x^2)+14]
=8
当且仅当x=1或-1时取等号