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

(- 2, 3)

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(2, 3)

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(- 4, 7)

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(4, 7)

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Solution

The correct option is A $(- 4, 7)$
Let $α = 2 + i \sqrt{3}$

Then, $β = 2 - i \sqrt{3}$

Therefore,
$α + β = 4$

$α β = (2 + i \sqrt{3}) (2 - i \sqrt{3}) = 4 + 3 = 7$

Therefore, equation is $x^{2} - 4 x + 7 = 0$

Therefore, $p = - 4$ and $q = 7$

Hence, option C is correct.

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Similar questions

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

If $(2 + i \sqrt{3})$ is a root of the equation $x^{2} + p x + q = 0$ where $p$ and $q$ are real, then find $(p, q)$ .

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

i \sqrt{3}

x^{2} + p x + q = 0

2 + i \sqrt{3}

x^{2} + p x + q = 0

p, q \in R

p + q

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