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JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Complex Numbers

If

α + i β

and

γ + i δ

are the roots of

x^{2} - (3 - 2 i) x - (2 i - 2) = 0, i = - 1

, then

α γ + β δ

is equal to :

(A)6
(B)2✓ Correct
(C)-2
(D)-6

Answer: (B)

Solution

For

x^{2} - (3 - 2 i) x - (2 i - 2) = 0

, the discriminant is

(3 - 2 i)^{2} + 4 (2 i - 2) = (9 - 12 i - 4) + (8 i - 8) = - 3 - 4 i

Writing

- 3 - 4 i = a + bi

,

a^{2} - b^{2} = - 3

,

2 ab = - 4

;

a = 1, b = - 2

works since

(1 - 2 i)^{2} = 1 - 4 i - 4 = - 3 - 4 i

.

x = \frac{( 3 - 2 i ) \pm ( 1 - 2 i )}{2}

gives

x = 2 - 2 i

and

x = 1

.

So

α + i β = 2 - 2 i

and

γ + i δ = 1

, i.e.

α = 2, β = - 2, γ = 1, δ = 0

.

α γ + β δ = 2 (1) + (- 2) (0) = 2

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Single correct option

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