Show that 2i and 1-iare both solutions of the equation x^(2)-(1+i)x+(2+2i)=0 but that their complex conjugates -2i and 1+i are not.

razdiral3m

razdiral3m

Answered question

2022-09-04

Show that 2i and 1-iare both solutions of the equation x 2 ( 1 + i ) x + ( 2 + 2 i ) = 0 but that their complex conjugates -2i and 1+i are not.

Answer & Explanation

eggma84rd

eggma84rd

Beginner2022-09-05Added 11 answers

we get:
x 2 ( 1 + i ) x + ( 2 + 2 i ) = 0
Using the quadratic formula, we then get:
x = ( 1 + i ) ± ( 1 + i ) 2 4 ( 1 ) ( 2 + 2 i ) 2 ( 1 )
= 1 + i ± 1 + 2 i 1 8 8 i 2 = 1 + i ± 8 6 i 2
= 1 + i ± ( 1 3 i ) 2
This means that:
x 1 = 1 i , x 2 = 2 i
Therefore, 2i and 1-i are solutions to the given equation, and theonly ones (because of the fundamental theorem of algebra).

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