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If $2+i \sqrt{3}$ is a root of the equation $x^2+p x+q=0$, where $p$ and $q$ are real, then $(p, q)=$
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$(-4,7)$
Since $2+i \sqrt{3}$ is a root, therefore $2-i \sqrt{3}$ will be other root. Now, sum of the roots $=4=-p$ and product of roots $=7=q$. Hence $(p, q)=(-4,7)$
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