(4+3i)x^2+mx+4-3i=0有实根,则(4+3i)x^2+mx+4-3i=(4x^2+mx+4)+(3x^2-3)i=0
所以4x^2+mx+4=0,3x^2-3=0,解得x=1或x=-1,代入4x^2+mx+4=0求得m=8,m=-8那原方程解怎么算？将m=8代入原方程得(4+3i)x^2+8X+4-3i=0(4x^2+8x+4)+(3x^2-3)i=0因有实根x,则4x^2+8x+4=0 x^2+2x+1=0x=-1m=-8时,同上解得x=1其实在求m时,3x^2-3=0,解得x=1或x=-1,这就是原方程的解