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

Question

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

experimentx · Answer

?? of a polynomial??

experimentx · Answer

complex roots always come in pairs!!

anonymous · Answer

sorry plzz  look the question now....

experimentx · Answer

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

experimentx · Answer

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

anonymous · Answer

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

anonymous · Answer

other is -i

experimentx · Answer

I think the same!!

anonymous · Answer

yes the other root is -i how?

anonymous · Answer

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

experimentx · Answer

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

experimentx · Answer

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

anonymous · Answer

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

anonymous · Answer

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.

experimentx · Answer

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!!

experimentx · Answer

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

anonymous · Answer

The answer is -i

anonymous · Answer

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