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

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\frac{- 1 - i}{2}

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i

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2 i

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Solution

The correct option is D $2 i$
$x^{2} + (1 - 3 i) x - 2 (1 + i) = 0$
One root $= - 1 + i$
Sum of roots $= 3 i - 1$
$2^{n d}$ root $= x$
$- 1 + i + x = 3 i - 1$
$x = 3 i - 1 + 1 - i$
$x = 2 i$
$∴$ other root $= 2 i$

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