Complex Numbers – Quadratic Equation – JEE Main 28 Jan 2025 Shift 2

Solution:

Given quadratic equation:
$$x^{2}-(3-2i)x-(2i-2)=0$$
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$:
$$x = \frac{(3-2i) \pm \sqrt{(3-2i)^{2} – 4(1)(-(2i-2))}}{2(1)}$$

Calculate the discriminant ($D$):
$$D = (3-2i)^{2} + 4(2i-2)$$
$$D = (9 – 4 – 12i) + 8i – 8$$
$$D = 5 – 12i + 8i – 8$$
$$D = -3 – 4i$$

We need to find $\sqrt{-3-4i}$. Let $\sqrt{-3-4i} = x+iy$.
Note that $(1-2i)^2 = 1 – 4 – 4i = -3-4i$.
So, $\sqrt{D} = \pm(1-2i)$.

Substitute $\sqrt{D}$ back into the roots equation:
$$x = \frac{(3-2i) \pm (1-2i)}{2}$$

Case 1 (Positive sign):
$$x_1 = \frac{3-2i + 1-2i}{2} = \frac{4-4i}{2} = 2-2i$$
Comparing with $\alpha+i\beta$, we get $\alpha = 2, \beta = -2$.

Case 2 (Negative sign):
$$x_2 = \frac{3-2i – (1-2i)}{2} = \frac{3-2i-1+2i}{2} = \frac{2}{2} = 1$$
Comparing with $\gamma+i\delta$, we get $\gamma = 1, \delta = 0$.

Now, we find the value of $\alpha\gamma + \beta\delta$:
$$\alpha\gamma + \beta\delta = (2)(1) + (-2)(0)$$
$$= 2 + 0 = 2$$

Ans. (2)