[UPLIFTJEE]

If $\alpha+\mathrm{i} \beta$ and $\gamma+\mathrm{i} \delta$ aretherootsof $\mathrm{x}^{2}-(3-2 \mathrm{i}) \mathrm{x}-(2 \mathrm{i}-2)=0, \mathrm{i}=\sqrt{-1}$,then $\alpha \gamma+\beta \delta$ is equal to [JEE (Main) 28 Jan 2025 S II]
(A) -6
(B) 6
(C) -2
(D) 2

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[UPLIFTJEE]

Solution

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

\[ \begin{aligned} & x=\frac{(3-2 i) \pm \sqrt{(3-2 i)^{2}-4(1)(-(2 i-2))}}{2(1)}=\frac{3-2 i \pm \sqrt{-3-4 i}}{2} \\ & x=\frac{3-2 i \pm \sqrt{(1)^{2}+(2 i)^{2}-2(1)(2 i)}}{2}=\frac{3-2 i \pm(1-2 i)}{2} \Rightarrow x=2-2 i \text { or } 1+0 i \end{aligned} \]

According to question $\alpha+\beta \mathrm{i}=2-2 \mathrm{i}$ and $\gamma+\delta \mathrm{i}=1+\mathrm{i} .0$ we get $\alpha=2, \beta=-2, \gamma=1, \delta=0$ So, $\alpha \gamma+\beta \delta=2(1)+(-2)(0)=2$

[UPLIFTJEE]

Answer

D

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