JEE Main 2026 Apr 6 Shift 1, Mathematics Q4: Quadratic Equations

JEE Main2026Apr 6, Shift 1Mathematics

Let the set of all values of $k \in R$ such that the equation $z (\overline{z} + 2 + i) + k (2 + 3 i) = 0$ , $z \in C$ , has at least one solution, be the interval $[α, β]$ . Then $9 (α + β)$ is equal to:

A
$- 10$
B
$- 8$
C
$1013$
D
$813$

Solution

Write $z = x + i y$ so that $\overline{z} + 2 + i = (x + 2) + i (1 - y)$ . Multiplying:

$z (\overline{z} + 2 + i) = (x + i y) ((x + 2) + i (1 - y))$ .

Separating real and imaginary parts and adding $k (2 + 3 i)$ :

Real: $x (x + 2) - y (1 - y) + 2 k = 0 \Rightarrow x^{2} + y^{2} + 2 x - y + 2 k = 0$ .

Imaginary: $x (1 - y) + y (x + 2) + 3 k = 0 \Rightarrow x + 2 y + 3 k = 0$ .

From the imaginary equation, $x = - 2 y - 3 k$ . Substituting into the real one and simplifying yields the quadratic in $y$ :

$5 y^{2} + (12 k - 5) y + 9 k^{2} - 4 k = 0$ .

A real $y$ requires non-negative discriminant: $(12 k - 5)^{2} - 20 (9 k^{2} - 4 k) \geq 0$ , which reduces to $36 k^{2} + 40 k - 25 \leq 0$ .

The roots of $36 k^{2} + 40 k - 25 = 0$ are $k = \frac{- 10 \pm 5 13}{18}$ , so by Vieta's $α + β = - \frac{40}{36} = - \frac{10}{9}$ .

$∴ 9 (α + β) = - 10$ .

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JEE Main 2026 Apr 6 Shift 1, Mathematics Q4: Quadratic Equations

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