If $\alpha + i\beta$ and $\gamma + i\delta$ are the roots of $x^2 - (3 - 2i)x - (2i - 2) = 0$, $i = \sqrt{-1}$, then $\alpha \gamma + \beta \delta$ is equal to:
Solution
<p>$$\begin{aligned}
& x^2-(3-2 i) x-(2 i-2)=0 \\
& x=\frac{(3-2 i) \pm \sqrt{(3-2 i)^2-4(1)(-(2 i-2))}}{2(1)} \\
& ==\frac{(3-2 i) \pm \sqrt{9-4-12 i+8 i-8}}{2} \\
& ==\frac{3-2 i \pm \sqrt{-3-4 i}}{2} \\
& =\frac{3-2 i \pm \sqrt{(1)^2+(2 i)^2-2(1)(2 i)}}{2} \\
& =\frac{3-2 \mathrm{i} \pm(1-2 \mathrm{i})}{2} \\
& \Rightarrow \frac{3-2 \mathrm{i}+1-2 \mathrm{i}}{2} \text { or } \frac{3-2 \mathrm{i}-1+2 \mathrm{i}}{2} \\
& \Rightarrow 2-2 \mathrm{i} \text { or } 1+0 \mathrm{i} \\
& \text { So } \alpha \gamma+\beta \delta=2(1)+(-2)(0)=2
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Complex Numbers and Quadratic Equations · Topic: Complex Numbers and Argand Plane
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