Theory of Equations 1 Question 40
41. If $2+i \sqrt{3}$ is a root of the equation $x^{2}+p x+q=0$, where $p$ and $q$ are real, then $(p, q)=(\ldots, \ldots)$.
(1982, 2M)
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Solution:
- If $2+i \sqrt{3}$ is one of the root of $x^{2}+p x+q=0$. Then, other root is $2-i \sqrt{3}$.
$$ \begin{aligned} \Rightarrow & & -p & =2+i \sqrt{3}+2-i \sqrt{3}=4 \\ \text { and } & & q & =(2+i \sqrt{3})(2-i \sqrt{3})=7 \\ \Rightarrow & & (p, q) & =(-4,7) \end{aligned} $$
