(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=0
x=-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...
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