A CUBIC EQUATION f(x) = 0 has a real root x and two complex roots a+ib and a-ib these points are represented as pts.A,B,C respectively in argand plane. SHOW THAT THE ROOTS OF THE DERIVED EQUATION f'(x) = 0 (i.e DERIVATIVE OF f(x) ) ARE COMPLEX. IF A FALLS ON ANY ONE OF THE TWO EQUILATERAL TRIANGLES DESCRIBED ON BASE BC
OD = a OA = x then AD = x - a
DC = b => tan 30 = b/(x-a) => x-a = \/3— b
The cubic eqn. is f(z) = z3 - (2a + x)z2 + (a2+b2+2ax)z - x(a2+b2) = 0
=>f ' (z) = 3z2 - 2(2a +x)z + a2+b2+2ax
D = Discriminant = {2(2a +x)}2 - 4(3)(a2+b2+2ax) = 4{(2a +x)2 - 3(a2+b2+2ax)}=4{4a2 + x2 + 4ax - 3a2-3b2-6ax}= 4(x2 - 2ax + a2 -3b2) = 4{(x-a)2 - 3b2}= 4(3b2- 3b2) = 0 Since D = 0 f ' (z) has only one root.