Exams Joint Entrance Examination Mathematics Complex Numbers Question 26 Question 26 of 455 Medium MCQ
If α + i β \alpha +i\beta α + i β and γ + i δ \gamma +i\delta γ + i δ are the roots of x 2 − ( 3 − 2 i ) x − ( 2 i − 2 ) = 0 {x}^{2}-(3-2i)x-(2i-2)=0 x 2 − ( 3 − 2 i ) x − ( 2 i − 2 ) = 0 , i = − 1 i=\sqrt{-1} i = − 1 , then α γ + β δ \alpha \gamma +\beta \delta α γ + β δ is equal to:
Solutions · 1 Post your solution Official solution
\[ x 2 − ( 3 − 2 i ) x − ( 2 i − 2 ) = 0 x = ( 3 − 2 i ) ± ( 3 − 2 i ) 2 − 4 ( 1 ) ( − ( 2 i − 2 ) ) 2 ( 1 ) = = ( 3 − 2 i ) ± 9 − 4 − 12 i + 8 i − 8 2 = = 3 − 2 i ± − 3 − 4 i 2 = 3 − 2 i ± ( 1 ) 2 + ( 2 i ) 2 − 2 ( 1 ) ( 2 i ) 2 = 3 − 2 i ± ( 1 − 2 i ) 2 ⇒ 3 − 2 i + 1 − 2 i 2 or 3 − 2 i − 1 + 2 i 2 ⇒ 2 − 2 i or 1 + 0 i So α γ + β δ = 2 ( 1 ) + ( − 2 ) ( 0 ) = 2 \begin{aligned}
& x^2-(3-2 i) x-(2 i-2)=0 \\
& x=\frac{(3-2 i) \pm \sqrt{(3-2 i)^2-4(1)(-(2 i-2))}}{2(1)} \\
& ==\frac{(3-2 i) \pm \sqrt{9-4-12 i+8 i-8}}{2} \\
& ==\frac{3-2 i \pm \sqrt{-3-4 i}}{2} \\
& =\frac{3-2 i \pm \sqrt{(1)^2+(2 i)^2-2(1)(2 i)}}{2} \\
& =\frac{3-2 \mathrm{i} \pm(1-2 \mathrm{i})}{2} \\
& \Rightarrow \frac{3-2 \mathrm{i}+1-2 \mathrm{i}}{2} \text { or } \frac{3-2 \mathrm{i}-1+2 \mathrm{i}}{2} \\
& \Rightarrow 2-2 \mathrm{i} \text { or } 1+0 \mathrm{i} \\
& \text { So } \alpha \gamma+\beta \delta=2(1)+(-2)(0)=2
\end{aligned} x 2 − ( 3 − 2 i ) x − ( 2 i − 2 ) = 0 x = 2 ( 1 ) ( 3 − 2 i ) ± ( 3 − 2 i ) 2 − 4 ( 1 ) ( − ( 2 i − 2 )) == 2 ( 3 − 2 i ) ± 9 − 4 − 12 i + 8 i − 8 == 2 3 − 2 i ± − 3 − 4 i = 2 3 − 2 i ± ( 1 ) 2 + ( 2 i ) 2 − 2 ( 1 ) ( 2 i ) = 2 3 − 2 i ± ( 1 − 2 i ) ⇒ 2 3 − 2 i + 1 − 2 i or 2 3 − 2 i − 1 + 2 i ⇒ 2 − 2 i or 1 + 0 i So α γ + β δ = 2 ( 1 ) + ( − 2 ) ( 0 ) = 2 \] Answer the question to unlock