Question

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

• A. (-4, 7)
• B. (4, -7)
• C. (-7,4)
• D. (4,7)

Answer 1

The correct option is A (-4, 7)

$(2+\sqrt{3i}{)}^2+p(2+\sqrt{3i})+q=0$

$4-3+4\sqrt{3}i+2p+\sqrt{3}pi+q=0$

$\Rightarrow (1+2p+q)+(4\sqrt{3}+\sqrt{3}p)i=0$

$\Rightarrow 2p+q=-1$; $p=-4$

$\Rightarrow q=-1+8=7$

$\Rightarrow$ (A).