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 :

Mathematics · Quadratic Equations

JEE Main 2025 — 28 January, Evening Shift — Question 4

If $\alpha+i \beta$ and $\gamma+i \delta$ are the roots of $x^{2}-(3-2 i) x-(2 i-2)=0, i=\sqrt{-1}$ , then $\alpha \gamma+\beta \delta$ is

equal to :

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

$x=\frac{(3-2 i) \pm \sqrt{(3-2 i)^{2}-4(1)(-(2 i-2))}}{2(1)}$

$==\frac{(3-2 \mathrm{i}) \pm \sqrt{9-4-12 \mathrm{i}+8 \mathrm{i}-8}}{2}$

$==\frac{3-2 \mathrm{i} \pm \sqrt{-3-4 \mathrm{i}}}{2}$

$=\frac{3-2 i \pm \sqrt{(1)^{2}+(2 i)^{2}-2(1)(2 i)}}{2}$

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

$\Rightarrow \frac{3-2 \mathrm{i}+1-2 \mathrm{i}}{2}$ or $\frac{3-2 \mathrm{i}-1+2 \mathrm{i}}{2}$

$\Rightarrow 2-2 \mathrm{i}$ or $1+0 \mathrm{i}$

So $\alpha \gamma+\beta \delta=2(1)+(-2)(0)=2$

Answer key and solution verified before publishing.

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