If 2 + $i \sqrt{3}$ is a root of $x^{2} + p x + q = 0$ , where p,q are real, then (p,q) is equal to

(-4, 7)

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(4, -7)

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(-7,4)

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(4,7)

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Solution

The correct option is A (-4, 7)
$(2 + \sqrt{3 i})^{2} + p (2 + \sqrt{3 i}) + q = 0$
$4 - 3 + 4 \sqrt{3} i + 2 p + \sqrt{3} p i + q = 0$
$\Rightarrow (1 + 2 p + q) + (4 \sqrt{3} + \sqrt{3} p) i = 0$
$\Rightarrow 2 p + q = - 1$ ; $p = - 4$
$\Rightarrow q = - 1 + 8 = 7$
$\Rightarrow$ (A).

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