Question

已知α,β为实系数二次方程ax^2+bx+c=0的两根,α为虚数,α^2/β属于实数,求α/β

Answer 1

由题意=>α,β共轭虚根x+yi ,x-yi

=> (x+yi)^2 /(x-yi) =(x^2-y^2+2xyi)/(x-yi) =(x^2-y^2+2xyi)(x+yi)/(x-yi)(x+yi)

=(x^2-y^2+2xyi)(x+yi)/(x^2+y^2)实数

=>分子虚部 2x^2y+(x^2-y^2)y=0=y(2x^2+x^2-y^2)=y(3x^2-y^2)

=> y=0 (不合) or 3x^2=y^2 => y=±(√3)x

=> x+yi=x+(√3)xi =x(1+√3 i) ;x-yi=x(1-√3 i) or x+yi=x(1-√3 i) ;x-yi=x(1+√3 i)

=> α/β=(1+√3 i)/(1-√3 i) or (1-√3 i)/(1+√3 i)

=(-1+√3 i)/2 or(-1-√3 i)/2 .ans