Solve x4+4x3+5x2+2x−2=0, given that −1+−1 is a root.

− 1 + i , − 1 − i , − 1 + 2 , 1 + 2

− 1 + i , − 1 − i , − 1 + 2 , 1 + 2

− 1 + i , − 1 − i , − 1 + 2 , − 1 − 2

− 1 + i , − 1 − i , − 1 + 2 , 1 + 2

1 + i , 1 − i , − 1 + 2 , − 1 − 2

1 + i , − 1 − i , − 1 + 2 , − 1 + 2

Book Online Demo

Check Your IQ

Try Test

Courses

Offline Centres

Solve $x^{4} + 4 x^{3} + 5 x^{2} + 2 x - 2 = 0$ , given that $- 1 + \sqrt{- 1}$ is a root.

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$- 1 + i, - 1 - i, - 1 + \sqrt{2}, - 1 - \sqrt{2}$

$- 1 + i, - 1 - i, - 1 + \sqrt{2}, 1 + \sqrt{2}$

$1 + i, 1 - i, - 1 + \sqrt{2}, - 1 - \sqrt{2}$

$1 + i, - 1 - i, - 1 + \sqrt{2}, - 1 + \sqrt{2}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

One root be $- 1 + i$ then the other root be $- 1 - i$

$\begin{array}{l} x^{2} - (- 2) x + (2) = 0 \\ x^{2} + 2 x + 2 = 0 \end{array}$

Solve $x^{2} + 2 x - 1$ then we will get remaining two roots

$\begin{array}{l} x^{2} + 2 x - 1 = 0 \\ x = \frac{- 2 \pm \sqrt{4 + 4}}{2} = \frac{- 2 \pm 2 \sqrt{2}}{2} = - 1 \pm \sqrt{2} \end{array}$

$∴$ Roots of the given equation are $- 1 + i, - 1 - i, - 1 + \sqrt{2}, - 1 - \sqrt{2}$

Watch 3-min video & get full concept clarity