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

Correct answer: D. 2 aligned & x^2-(3-2 i) x-(2 i-2)=0 & x= (3-2 i) ± (3-2 i)^2-4(1)(-(2 i-2)) 2(1) & == (3-2 i) ± 9-4-12 i+8 i-8 2 & == 3-2 i ± -3-4 i 2 & = 3-2 i ± (1)^2+(2 i)^2-2(1)(2 i) 2 & = 3-2 i ±(1-2 i) 2 & 3-2 i+1-2 i 2 or 3-2 i-1+2 i 2 & 2-2 i or 1+0 i & So α γ+β =2(1)+(-2)(0)=2 aligned

JEE Main 202528 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 28 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Complex Number

JEE Main 2025 · 28 January · Evening Shift · Mathematics · Complex Number

If α+i β and γ+i are the roots of x^2-(3-2 i) x-(2 i-2)=0, i= -1 ,...

\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 :

Options

A $-2$
B $6$
C $-6$
D $2$

Correct answer

D. $2$

Step-by-step solution

\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} \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}

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