Complex Numbers And Quadratic Equations question 694
Question: Let $ \lambda \in \mathbf{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 arg and plane. Then $ \lambda $ is
Options:
A) 1
B) $ \frac{2}{3} $
C) 2
D) $ -1 $
Show Answer
Answer:
Correct Answer: B
Solution:
For the nonreal roots of the equation $ 2z^{2}+2z+\lambda =0 $ … (i) $ discriminant <0. $ That is $ 4-8\lambda <0\Rightarrow \lambda >\frac{1}{2} $ … (ii) Let the roots of (i) be $ {z_1} \And z_2 $ Then $ {z_1}+z_2=-\frac{2}{2}=-1,z_1z_2=\frac{\lambda }{2} $ $ {z_1}andz_2 $ with origin form equilateral triangle if $ {z^{2}}+z_2^{2}-z_1z_2=0 $
$ \Rightarrow {{(z_1+z_2)}^{2}}=3z_1z_2 $
$ \Rightarrow {{(-1)}^{2}}=3\frac{\lambda }{2}\Rightarrow \lambda =\frac{2}{3} $ $ \lambda =\frac{2}{3}( >\frac{1}{2} ) $ satisfies the condition (ii). Hence it is the required result.
BETA
