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The equation of the smallest degree with real coefficients having $1 + i$ as one of the roots is

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$x^{2} + x + 1 = 0$

$x^{2} - 2 x + 2 = 0$

$x^{2} + 2 x + 2 = 0$

$x^{2} + 2 x - 2 = 0$

answer is B.

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Detailed Solution

Let $x = 1 + i \Leftrightarrow$ $x - 1 = i$

Squaring on both sides,

we get

$\Leftrightarrow$ ${(x - 1)}^{2} = i^{2} \Rightarrow x^{2} - 2 x + 1 = - 1$

$\Leftrightarrow$ $x^{2} - 2 x + 2 = 0.$

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