Question

Let $\lambda ϵR$. If the origin and the nonreal roots of $2z^2+2z+\lambda =0$ form the three vertices of an equilateral triangle in the Argrand plane then$\lambda$ is

• A. $1$
• B. $\frac{2}{3}$
• C. $2$
• D. $-1$

Answer 1

Answer: (B)

$\frac{2}{3}$

$z_1{z}_2=\frac{\lambda }{2}$ and
$z_1+z_2=-1$
Now since 0,$z_1,z_2$ form an equilateral triangle.
$z_1\left(z_2\right)+z_2\left(0\right)+z_1\left(0\right)=z_1^2+z_2^2+0^2$
$z_1^2+z_2^2=z_2{z}_1$
$(z_1+z_2{)}^2-2z_1{z}_2=z_1{z}_2$
$1=3z_1{z}_2$
$1=3\left(\frac{\lambda }{2}\right)$
$\lambda =\frac{2}{3}$