If one root of the equation ix^2 -2(i+1)x +2-i=0 , is 2-i then the other root is ?
?? of a polynomial??
complex roots always come in pairs!!
sorry plzz look the question now....
sorry .. i didn't see 2-i there
(x - (2-i))(x - b) = (ix^2 -2(i+1)x +2-i)/i <-- solve from here
[x = 2-I], [x = -I]
other is -i
I think the same!!
yes the other root is -i how?
use quadratic formula and i think discriminant will be sqrt (-1) somehow
b(2-i) = (2-i)/i = 1/i = -i
sorry, b = 1/i = -i check the last portion (without variable of the equation)
b^2-4ac......
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.
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!!
i mean the coefficients of those two equations!! since they are same equations ... they must have same coefficients!!
The answer is -i
|dw:1338088880980:dw|
Join our real-time social learning platform and learn together with your friends!