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Equilateral Triangle Roots of 2z²+2z+k=0 Find k | JEE

JEE Maths question with a full step-by-step solution.

Question

One vertex of an equilateral triangle is at the origin and the other two vertices are the roots of

2z^2 + 2z + k = 0

. Then the value of

k

1

\dfrac{1}{3}

\dfrac{2}{3}

correct

\dfrac{1}{2}

Solution

Step 1: Let the two roots be

z_1

and

z_2

. By Vieta's formulas for

2z^2 + 2z + k = 0

(divide through by

2

to get

z^2 + z + \tfrac{k}{2} = 0

z_1 + z_2 = -1, \qquad z_1 z_2 = \frac{k}{2}

Step 2: Apply the equilateral-triangle condition. Three points

z_1, z_2, z_3

form an equilateral triangle if and only if:

z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1

Step 3: Substitute the third vertex

z_3 = 0

(the origin):

z_1^2 + z_2^2 + 0 = z_1 z_2 + 0 + 0 \implies z_1^2 + z_2^2 = z_1 z_2

Step 4: Express

z_1^2 + z_2^2

using the identity

z_1^2 + z_2^2 = (z_1 + z_2)^2 - 2z_1 z_2

(z_1 + z_2)^2 - 2z_1 z_2 = z_1 z_2 \implies (z_1 + z_2)^2 = 3z_1 z_2

Step 5: Substitute the Vieta values:

(-1)^2 = 3\cdot\frac{k}{2} \implies 1 = \frac{3k}{2} \implies k = \frac{2}{3}

Correct answer: (3)

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