JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations
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 :
- A \(-2\)
- B \(6\)
- C \(-6\)
- D \(2\)
Answer & Solution
Correct Answer
(D) \(2\)
Step-by-step Solution
Detailed explanation
\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…
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