Question

If $2+i\sqrt{3}$ is a root of the equation $x^2+px+q=0$, where $p,q\in R$, then find $p+q$.

Answer 1

According to the rule of conjugate pairs,

The roots of $x^2+px+q=0$ are

$[2+i\sqrt{3}]$ and $[2-i\sqrt{3}]$

Now, for $x^2+px+q=0;$

Sum of roots $=-p$

$\Rightarrow$ $[2+i\sqrt{(}3)]+[2-i\sqrt{(}3\left)\right]$ $=-p$

$\Rightarrow$ $4=-p$

$\Rightarrow$ $p=-4$

Similarly,

Product of roots $=q$

$\Rightarrow$ $[2+i\sqrt{(}3)]\times [2-i\sqrt{(}3\left)\right]=q$

$\Rightarrow$ $(2{)}^2-(i\sqrt{(}3){)}^2=q$ using $\left(\right(a+b\left)\right(a-b)=a^2-b^2)$

$\Rightarrow$ $4-i^2.3=q$

$\Rightarrow$ $4-\left(3\right)=q$

$\Rightarrow$ $q=7$

So, $p=-4$ and $q=7$

Ordered pair is $(-4,7)$

$p+q=-4+7=3$