Question

已知实数系方程x^2+px+q=0的两根在复平面上所对应的两点与原点恰为一个正三角形的

Answer 1

设两根为a+bi和a-bi

对应两点长度为2b

所以a^2+b^2 = 4b^2 =>a^2=3b^2

p = -2a

a^2+b^2 = q

p^2 = 4a^2 = 12b^2

q = 4b^2

所以p^2 = 3q

(2)

a^2+b^2 = 4 = 4b^2 = q

p = Sqrt[3q]=Sqrt[12]