The origin and the roots of the equation x2+ax+b = 0 form an equilateral triangle, if :
a2=3b
x2+ax+b = 0 ⇒−a±√a2−4b2
If a2−4b≤0 then the three points, 0 and the two roots are real and so collinear. Therefore, considering a2−4b<0, we have roots as −a±√4b−a22
Let A = −a2+i√4b−a22, B = −a2−i√4b−a22
OA = √a24+4b−a24 = √b = OB
AB = √(4b−a2). Since OA = OB = AB,
we get 4b−a2 = b i.e., a2 = 3b