If alpha+i beta and gamma+i delta are the roo | vedclass

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

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(B) Given the quadratic equation $x^2-(3-2 i) x-(2 i-2)=0$. \ Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$: \ $x = \frac{(3-2 i) \pm \sqrt{(3-2 i)^2 - 4(1)(-(2 i-2))}}{2}$ \ $x = \frac{(3-2 i) \pm \sqrt{9 - 4 - 12i + 8i - 8}}{2}$ \ $x = \frac{(3-2 i) \pm \sqrt{-3 - 4i}}{2}$ \ Since $-3-4i = 1^2 + (2i)^2 - 2(1)(2i) = (1-2i)^2$,we have: \ $x = \frac{(3-2 i) \pm (1-2 i)}{2}$ \ Case $1$: $x = \frac{3-2i + 1-2i}{2} = \frac{4-4i}{2} = 2-2i$. Here $\alpha=2, \beta=-2$. \ Case $2$: $x = \frac{3-2i - (1-2i)}{2} = \frac{2}{2} = 1+0i$. Here $\gamma=1, \delta=0$. \ Thus,$\alpha \gamma + \beta \delta = (2)(1) + (-2)(0) = 2+0 = 2$.

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

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