If-x-1-i-is-a-root-of-the-equation-x-3-ix-1-i-0-then-the-other-real-root-is-

Question Number 49460 by Pk1167156@gmail.com last updated on 07/Dec/18

If x = 1 + i is a root of the equation

x^{3} - i x + 1 - i = 0, then the other real

root is

Commented by Kunal12588 last updated on 07/Dec/18

i d o i t w i t h n o r m a l r e a l w a y . s o i t

s h o u l d b e w r o n g .

x^{3} - i x + 1 - i = (x - 1 - i) (x + i) (x + 1)

r o o t s a r e

x_{1} = 1 + i, x_{2} = - i, x_{3} = - 1 . s o r e a l r o o t = - 1 .

p l e a s e i n f o r m i f i t i s w r o n g .

Commented by Abdo msup. last updated on 08/Dec/18

(x - 1 - i) (x + i) (x + 1) = (x - 1 - i) (x^{2} + x + i x + i)

= (x - 1 - i) (x^{2} + (1 + i) x + i)

= x^{3} + (1 + i) x^{2} + i x - x^{2} - (1 + i) x - i - {i x}^{2} - i (1 + i) x + 1

x^{3} - x - i x + x + 1 - i = x^{3} - i x + 1 - i s o y o u r a n s w e r i s

c o r r e c t s i r .