Question

Let $\lambda \in R$ 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 1

Answer: (B)

$\frac{2}{3}$

The equation $\Rightarrow 2z^2+2z+⋋=0$

$\Rightarrow z^2+z+\frac{⋋}{2}=0$

Let $z_1$ and $z_2$ be the two roots of the equation

$z_1+z_2=-1$

$z_1{z}_2=\frac{⋋}{2}$

$so,z_2=z_1\left(\cos 60°+\sin 60°\right)$

$z_1(\frac{1}{2}+\frac{\sqrt{3}}{2}i)$

so, $z_1-z_2=\sqrt{3i}+z_1$

$\therefore {(2z_2^2-z_1)}^2=-3z_1^2$

$(z_1^2-z_2^2)=z_1{z}_2$

$1-\left(\frac{⋋}{2}\right).2=\left(\frac{⋋}{z}\right)$

$⋋=\frac{2}{3}$