Question

If $2+i\sqrt{3}$ is a root of $x^2+px+q=0$, then the values of $p$ and $q$ are respectively

• A. $p=4;q=7$
• B. $p=4;q=3$
• C. $p=4;q=-7$
• D. $p=-4;q=7$

Answer 1

The correct option is D $p=-4;q=7$

We know that if a quadratic equation has non-real (imaginary) roots, its roots always exist in conjugate pairs. for eg., $(a+ib)$ and $(a-ib)$

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 the roots $=-p$

$\therefore$ $2+i\sqrt{3}+2-i\sqrt{3}=-p$

$\therefore$ $4=-p$

$\therefore$ $p=-4$

Similarly, product of the rots $=q$

$\therefore$ $(2+i\sqrt{3})\times (2-\sqrt{3})=q$

$\therefore$ $(2{)}^2-(i.\sqrt{2}{)}^2=q$

$\therefore$ $4-3i^2=q$

$\therefore$ $4+3=q$

$\therefore$ $q=7$