JEE Main 2025 Jan 28 Shift 2, Mathematics Q4: Quadratic Equations

JEE Main2025Jan 28, Shift 2Mathematics

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
$2$
C
$- 2$
D
$- 6$

Solution

Apply the quadratic formula. The discriminant is

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

We need $- 3 - 4 i$ . Writing $- 3 - 4 i = (1 - 2 i)^{2}$ (since $(1 - 2 i)^{2} = 1 - 4 i + 4 i^{2} = - 3 - 4 i$ ), gives $- 3 - 4 i = \pm (1 - 2 i)$ .

Thus $x = \frac{( 3 - 2 i ) \pm ( 1 - 2 i )}{2}$ , giving the two roots $x_{1} = 2 - 2 i$ and $x_{2} = 1 + 0 i$ .

So $(α, β) = (2, - 2)$ and $(γ, δ) = (1, 0)$ , hence $α γ + β δ = 2 \cdot 1 + (- 2) \cdot 0 = 2$ .

Concept behind this question

Quadratic EquationsNotes, formulas and examples →

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2025 Jan 28 Shift 2, Mathematics Q4: Quadratic Equations

Concept behind this question

More previous year questions on Quadratic Equations

Practice more Mathematics