If one root of the equation $x^{2} + (1 - 3 i) x - 2 (1 + i) = 0$ is $- 1 + i$ , then the other root is

- 1 - i

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{(- 1 - i)}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

i

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 i

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is A $2 i$
Let another root of the equation
$x^{2} + (1 - 3 i) x - 2 (1 + i) = 0 i s α$
$∴ a + (- 1 + i) = - (1 - 3 i)$
$\Rightarrow a = - 1 + 3 i + 1 - i$
$= 2 i$
x2+(1−3i)x−2(1+i)=0isα

Suggest Corrections

Similar questions

x^{2} + (1 - 3 i) x - 2 (1 + i) = 0

- 1 + i,

x^{2} - i x - (1 + i) = 0

1 + i

i = \sqrt{- 1}

i x^{2} - 2 (i + 1) x + (2 - i) = 0

2 - i

x^{2} - (3 + 2 i) x + (1 + 3 i) = 0

1 + i

i = \sqrt{- 1}

x = 1 + i

x^{3} - i x + 1 - i = 0

Join BYJU'S Learning Program