If –1 + i is a root of $x^{4} + 4 x^{3} + 5 x^{2} + 2 x + k = 0$ then its real roots are

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–1, –1

$1 \pm \sqrt{2}$

$- 1 \pm \sqrt{2}$

$- \frac{1}{2}, - \frac{3}{2}$

answer is D.

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Detailed Solution

$\begin{array}{l} I f - 1 + i i s r o o t o f x^{4} + 4 x^{3} + 5 x^{2} + 2 x + k = 0 \\ T h e n - 1 - i i s a l s o r o o t \\ ∴ T h e E q u a t i o n i s x^{2} + 2 x + 2 = 0 \\ ∴ x^{4} + 4 x^{3} + 5 x^{2} + 2 x + k i s d i v i d e d b y x^{2} + 2 x + 2 \\ B y s y n t h e t i c D i v i s i o n \end{array}$

$\begin{array}{l} ∴ k + 2 = 0, k = - 2 \\ ∴ x^{4} + 4 x^{3} + 5 x^{2} + 2 x - 2 = (x^{2} + 2 x + 2) (x^{2} + 2 x - 1) \\ x^{2} + 2 x - 1 r o o t s a r e \\ x = \frac{- 2 \pm \sqrt{4 + 4}}{2} \Rightarrow \frac{2 \pm 2 \sqrt{2}}{2} \Rightarrow 1 \pm \sqrt{2} \end{array}$