Question

One vertex of an equilateral triangle is at the origin and the other two vertices are given by 2z² + 2z + k = 0, then k is –

• A. 2/3
• B. 1
• C. 2
• D. –1

Answer 1

Answer: (A)

2/3

Answer 2

Sol. z = $\frac{- 2 \pm \sqrt{4 - 8k}}{4}$

Since z is a complex number, we must have 4 – 8k = –ive or 2k – 1 > 0 or k > 1/2 … (1)

Hence the roots are –$\frac{1}{2}$ ± $\frac { \mathrm { i } } { 2 }$ $\sqrt{2k - 1}$

\ Vertices A, B, C are

(0, 0), $\left( - \frac{1}{2} + \frac{i}{2}\sqrt{2k - 1} \right)$, $\left( - \frac{1}{2}–\frac{i}{2}\sqrt{2k - 1} \right)$

Since the triangle is equilateral

\ |AB| = |BC| = |CA|

$\frac{1}{4} + \frac{1}{4}$ (2k – 1) = $\sqrt{2k - 1}$)²

or $\frac{1}{2}$k = 2k – 1 \ $\frac{3}{2}$k = 1

or k = $\frac{2}{3}$ and it is > $\frac{1}{2}$ as required by (1).