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**Hint:**We have a quadratic equation as: ${{x}^{2}}+px+q=0$ whose one root is $2+i\sqrt{3}$. So, we need to find the other root first. Then by using the sum of roots and product of roots formula, find the value of p and q.

**Complete step by step answer:**

As we know that, for a quadratic equation: $a{{x}^{2}}+bx+c=0$ , if one root is $\alpha +i\beta $ so the other root will be $\alpha -i\beta $

So, for the quadratic equation given in the question: ${{x}^{2}}+px+q=0$. Since one root is $2+i\sqrt{3}$. Therefore, the other root will be $2-i\sqrt{3}$

Now, we know that: for a quadratic equation: $a{{x}^{2}}+bx+c=0$

Sum of roots is: $-\dfrac{b}{a}$

Product of roots is: \[\dfrac{c}{a}\]

So, for the quadratic equation given in the question: ${{x}^{2}}+px+q=0$

Sum of roots is: $-\dfrac{p}{1}......(1)$

Product of roots is: \[\dfrac{q}{1}......(1)\]

Also, we know that the roots of the quadratic equation are: $2+i\sqrt{3}$ and $2-i\sqrt{3}$

So, we can write equation (1) and equation (2) as:

\[\begin{align}

& 2+i\sqrt{3}+2-i\sqrt{3}=-p......(3) \\

& \left( 2+i\sqrt{3} \right)\left( 2-i\sqrt{3} \right)=q......(4) \\

\end{align}\]

By solving equation (4) and equation (5) using $i=\sqrt{-1}$ , we get the value of p and q as:

$\begin{align}

& \Rightarrow 2+2=-p \\

& p=-4 \\

& \Rightarrow \left( 2\times 2 \right)+\left( 2\times \left( -i\sqrt{3} \right) \right)+\left( i\sqrt{3}\times 2 \right)+\left( i\sqrt{3}\times \left( -i\sqrt{3} \right) \right)=q \\

& 4-2i\sqrt{3}+2i\sqrt{3}+3=q \\

& q=7 \\

\end{align}$

So, $\left( p,q \right)=\left( -4,7 \right)$

**So, the correct answer is “Option A”.**

**Note:**While applying the identity for the sum of zeros and products of zeros, always take care of the negative sign in the sum of zeros. Some deliberately miss out on the use of negative signs in the formula and this gives you the wrong value. Also, it was given in the question, that the polynomial is quadratic. For a higher degree of the polynomial, the formula for the sum of zeros and product of zeros changes accordingly.

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