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  4. If 2+√5 i is a root of x2-p x+q=0 where p and q are real, then the ordered pair (p, q) is equal to

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Q. If $2+\sqrt{5} i$ is a root of $x^2-p x+q=0$ where $p$ and $q$ are real, then the ordered pair $(p, q)$ is equal to

Complex Numbers and Quadratic Equations

Solution:

As $p, q \in R$, other root of the equation is $2-i \sqrt{5}$.
Thus, $p=4, q=(2+\sqrt{5} i)(2-\sqrt{5} i)=9$