Let the set of all values of such that the equation , , has at least one solution, be the interval . Then is equal to : [2026]

Question

Let the set of all values of $k \in R$ such that the equation $z (\bar{z} + 2 + i) + k (2 + 3 i) = 0$ , $z \in C$ , has at least one solution, be the interval $[α, β]$ . Then $9 (α + β)$ is equal to : [2026]

Smart Achievers · Accepted Answer

(1)

We have, $z (\bar{z} + 2 + i) + k (2 + 3 i) = 0$

Let z = x + iy

Then, (x + iy)(x – iy + 2 + i) + k(2 + 3i) = 0

$\Rightarrow (x^{2} + y^{2} + 2 x - y + 2 k) + i (x + 2 y + 3 k) = 0$

On comparing real and imaginary parts, we get

$x^{2} + y^{2} + 2 x - y + 2 k = 0$ ... (i)

and x + 2y + 3k = 0 ... (ii)

$\Rightarrow x = - (2 y + 3 k)$

Put in (i), we get

${[- (2 y + 3 k)]}^{2} + y^{2} + [- 2 (2 y + 3 k)] - y + 2 k = 0$

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

$∵$ y is real

$∴ D \geq 0$

${(12 k - 5)}^{2} - 4 \times 5 (9 k^{2} - 4 k) \geq 0$

$\Rightarrow 36 k^{2} + 40 k - 25 \leq 0$

$\Rightarrow \frac{- 40 - 20 \sqrt{13}}{72} \leq k \leq \frac{- 40 + 20 \sqrt{13}}{72}$

$∴ α = \frac{- 40 - 20 \sqrt{13}}{72} and β = \frac{- 40 + 20 \sqrt{13}}{72}$

So, $9 (α + β) = 9 (- \frac{40}{72} \times 2) = - 10$

Solution

Q.
Let the set of all values of $k \in R$ such that the equation $z (\bar{z} + 2 + i) + k (2 + 3 i) = 0$ , $z \in C$ , has at least one solution, be the interval $[α, β]$ . Then $9 (α + β)$ is equal to : [2026]

1 –10

2 –8

3 $10 \sqrt{13}$

4 $8 \sqrt{13}$

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