Question
Let $$\lambda \in {\bf{R}}.$$ If the origin and the non real roots of $$2z^2 + 2z + \lambda = 0$$ form the three vertices of an equilateral triangle in the argand plane. Then $$\lambda$$ is
A.
$$1$$
B.
$$\frac{2}{3}$$
C.
$$2$$
D.
$$- 1$$
Answer :
$$\frac{2}{3}$$
Solution :
For the non-real roots of the equation $$2{z^2} + 2z + \lambda = 0\,\,\,\,\,.....\left( {\text{i}} \right)$$
$$\eqalign{
& {\text{discriminant}} < 0. \cr
& {\text{That}}\,{\text{is}}\,\,4 - 8\lambda < 0 \cr
& \Rightarrow \,\lambda > \frac{1}{2}\,\,\,\,\,.....\left( {{\text{ii}}} \right) \cr} $$
Let the roots of (i) be $${z_1}\,\,\& \,\,{z_2}$$
$${\text{Then}}\,\,{z_1} + {z_2} = - \frac{2}{2} = - 1,\,{z_1}{z_2} = \frac{\lambda }{2}$$
$$\eqalign{
& {z^2} + z_2^2 - {z_1}{z_2} = 0 \cr
& \Rightarrow \,{\left( {{z_1} + {z_2}} \right)^2} = 3{z_1}{z_2} \cr
& \Rightarrow \,{\left( { - 1} \right)^2} = 3\frac{\lambda }{2} \cr
& \Rightarrow \,\lambda = \frac{2}{3} \cr} $$
$$\lambda = \frac{2}{3}\left( { > \frac{1}{2}} \right)$$ satisfies the condition (ii).
Hence, it is the required result.