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OpenStudy (anonymous):

If one root of the equation ix^2 -2(i+1)x +2-i=0 , is 2-i then the other root is ?

OpenStudy (experimentx):

?? of a polynomial??

OpenStudy (experimentx):

complex roots always come in pairs!!

OpenStudy (anonymous):

sorry plzz look the question now....

OpenStudy (experimentx):

sorry .. i didn't see 2-i there

OpenStudy (experimentx):

(x - (2-i))(x - b) = (ix^2 -2(i+1)x +2-i)/i <-- solve from here

OpenStudy (anonymous):

[x = 2-I], [x = -I]

OpenStudy (anonymous):

other is -i

OpenStudy (experimentx):

I think the same!!

OpenStudy (anonymous):

yes the other root is -i how?

OpenStudy (anonymous):

use quadratic formula and i think discriminant will be sqrt (-1) somehow

OpenStudy (experimentx):

b(2-i) = (2-i)/i = 1/i = -i

OpenStudy (experimentx):

sorry, b = 1/i = -i check the last portion (without variable of the equation)

OpenStudy (anonymous):

b^2-4ac......

OpenStudy (anonymous):

well if i=sqrt(-1) then -i= -sqrt (-1) plug in -sqrt(-1) into x to see for your self or us the qudratic formula and see.

OpenStudy (experimentx):

let the other root be b \[ (x - (2-i))(x-b) = \frac{(ix^2 -2(i+1)x +(2-i))}{i} \] \[ x^2 -(2-i+b)x + b(2-i) = x^2 - \frac{2(i+1)}{i}x + \frac{2-i}{i}\] compare those two eqautions!!

OpenStudy (experimentx):

i mean the coefficients of those two equations!! since they are same equations ... they must have same coefficients!!

OpenStudy (anonymous):

The answer is -i

OpenStudy (anonymous):

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