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

Question

Dr. Abubacker · Accepted Answer

(a)
Since, $2 + i\sqrt{3}$ is a root of equation $x^{2} + px + q = 0$. Therefore, $2 - i\sqrt{3}$ will be other root.
Now, Sum of the roots $= \left( 2 + i\sqrt{3} \right) + (2 - i\sqrt{3}$ )$= - p$
$\Rightarrow 4 = - p$
Product of roots $= \left( 2 + i\sqrt{3} \right) + (2 - i\sqrt{3}) = q$
$\Rightarrow 7 = q$
Hence, $(p,q) = ( - 4,\ 7)$

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

Explanation

Reviewed by Dr. Abubacker

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