Question

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

Answer 1

?? of a polynomial??

Answer 2

complex roots always come in pairs!!

Answer 3

sorry plzz look the question now....

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

other is -i

Answer 8

I think the same!!

Answer 9

yes the other root is -i how?

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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.

Answer 15

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!!

Answer 16

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

Answer 17

The answer is -i