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 :

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Aicharya · Accepted Answer

$\begin{aligned} & x^2-(3-2 \mathrm{i}) \mathrm{x}-(2 \mathrm{i}-2)=0 \ & \mathrm{x}=\frac{(3-2 \mathrm{i}) \pm \sqrt{(3-2 \mathrm{i})^2-4(1)(-(2 \mathrm{i}-2))}}{2(1)} \ & ==\frac{(3-2 \mathrm{i}) \pm \sqrt{9-4-12 \mathrm{i}+8 \mathrm{i}-8}}{2} \ & ==\frac{3-2 \mathrm{i} \pm \sqrt{-3-4 \mathrm{i}}}{2} \ & =\frac{3-2 \mathrm{i} \pm \sqrt{(1)^2+(2 \mathrm{i})^2-2(1)(2 \mathrm{i})}}{2} \ & =\frac{3-2 \mathrm{i} \pm(1-2 \mathrm{i})}{2} \ & \Rightarrow \frac{3-2 \mathrm{i}+1-2 \mathrm{i}}{2} 	ext { or } \frac{3-2 \mathrm{i}-1+2 \mathrm{i}}{2} \ & \Rightarrow 2-2 \mathrm{i} 	ext { or } 1+0 \mathrm{i} \ & 	ext { So } \alpha \gamma+\beta \delta=2(1)+(-2)(0)=2\end{aligned}$

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…

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