lf $2 + i \sqrt{3}$ is one root of $x^{2} + p x + q = 0$ then $p, q =$
(Note: $p, q$ are real numbers)

4, 7

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1, 1

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- 4, 7

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4, 4

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Solution

The correct option is B $- 4, 7$
The coefficients are real. If one of the roots is non-real, the other root must be its conjugate.
Hence, $2 - i \sqrt{3}$ is also a root of the equation. Hence,
$s u m o f r o o t s = - p = 4$
$q = p r o d u c t o f r o o t s = 2^{2} + {\sqrt{3}}^{2} = 7$

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