如果方程(1+i)x²-2(a-i)x+5-3i=0(a∈R)有实数解,求a的值
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如果方程(1+i)x²-2(a-i)x+5-3i=0(a∈R)有实数解,求a的值
如果方程(1+i)x²-2(a-i)x+5-3i=0(a∈R)有实数解,求a的值
如果方程(1+i)x²-2(a-i)x+5-3i=0(a∈R)有实数解,求a的值
(x^2-2ax+5)+i(x^2+2x-3)=0
虚部为0-->x^2+2x-3=0-->(x-1)(x+3)=0--->x=1 or -3
实部为0-->x^2-2ax+5=0-->x=1:1-2a+5=0--> a=3
x=-3:9+6a+5=0-->a=-7/3
a=3 或-7/3.