Let f(x)=ax2+bx+c,a,b,c∈R. If f (x) takes real values for real values of x and non-real values for non-real values of x, then

Suppose a≠0. We rewrite f(x) as follows:f(x)=ax2+bax+ca=ax+b2a2+4ac−b24a2f−b2a+i=a−b2a+i+b2a2+4ac−b24a2 =a−1+4ac−b24a2, which is a real number This is against the hypothesis. Therefore, a = 0.