If α+iβ and γ+iδ are the roots of x2−(3−2i)x−(2i−2)=0,i=−1 ,then αγ+βδ is equal to :

x2−(3−2i)x−(2i−2)=0==(3−2i)±(3−2i)2−4(1)(−(2i−2))2(1)==3−2i±−3−4i2=3−2i±(1)2+(2i)2−2(1)(2i)2=3−2i±(1−2i)2⇒3−2i+1−2i2 or 3−2i−1+2i2⇒2−2i or 1+0i So αγ+βδ=2(1)+(−2)(0)=2

If α+iβ and γ+iδ are the roots of x2−(3−2i)x−(2i−2)=0,i=−1 ,then αγ+βδ is equal to :

x2−(3−2i)x−(2i−2)=0==(3−2i)±(3−2i)2−4(1)(−(2i−2))2(1)==3−2i±−3−4i2=3−2i±(1)2+(2i)2−2(1)(2i)2=3−2i±(1−2i)2⇒3−2i+1−2i2 or 3−2i−1+2i2⇒2−2i or 1+0i So αγ+βδ=2(1)+(−2)(0)=2

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If $α + i β and γ + i δ$ are the roots of $x^{2} - (3 - 2 i) x - (2 i - 2) = 0, i = \sqrt{- 1}$ ,then $α γ + β δ$ is equal to :

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Detailed Solution

$\begin{array}{l} x^{2} - (3 - 2 i) x - (2 i - 2) = 0 \\ == \frac{(3 - 2 i) \pm \sqrt{(3 - 2 i)^{2} - 4 (1) (- (2 i - 2))}}{2 (1)} \\ == \frac{3 - 2 i \pm \sqrt{- 3 - 4 i}}{2} \\ = \frac{3 - 2 i \pm \sqrt{(1)^{2} + (2 i)^{2} - 2 (1) (2 i)}}{2} \\ = \frac{3 - 2 i \pm (1 - 2 i)}{2} \\ \Rightarrow \frac{3 - 2 i + 1 - 2 i}{2} or \frac{3 - 2 i - 1 + 2 i}{2} \\ \Rightarrow 2 - 2 i or 1 + 0 i \\ So α γ + β δ = 2 (1) + (- 2) (0) = 2 \end{array}$