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)

Question

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

Professor PC Thomas Classes · Answer

Since $2+i \sqrt{3}$ is a root, $2-i \sqrt{3}$ is also a root\ $(2+i \sqrt{3})+(2-\mathrm{i} \sqrt{3})=-p \quad 	herefore p=-4$\ $(2+i \sqrt{3})+(2-\mathrm{i} \sqrt{3})=q \quad 	herefore q=7$\ $	herefore \quad(p, q)=(-4,7)$.\