已知复数Z为实系数一元二次方程ax^2+bx+1=0的根,且（1-3i）Z=（-2+i）Z+1-i

题目

已知复数Z为实系数一元二次方程ax^2+bx+1=0的根,且（1-3i）Z=（-2+i）Z+1-i

答案

(1-3i)Z = (-2+i)Z + 1 - i
[(1-3i)-(-2+i)]Z = 1-i
Z = (1-i)/(3-4i)
= (1-i)(3+4i)/(9+16)
= (7+i)/25
ax^2+bx+1 = 0
(7+i)/25是其中一个根,（7-i)/25为另外一根
(7+i)/25 + (7-i)/25 = -b/a
-b/a = 14/25 (1)
[(7+i)/25][(7-i)/25]= 1/a
a = 25/2
把 a = 25/2 带入 （1）
-b/(25/2) = 14/25
b = -7
所以
25x^2/2 - 7x + 1 = 0
25x^2 - 14x + 2 = 0