A) (-4, 7) |
B) (4,-7) |
C) (4, 7) |
D) (-4, -7) |
A) (-4, 7) |
Since, \( \Large 2+i\sqrt{3} \) is a root of equation \( \Large x^{2}+px+q=0 \) therefore, \( \Large 2-i\sqrt{3} \) will be other root. Now sum of the roots \( \Large 2+i\sqrt{3} \) + \( \Large 2-i\sqrt{3} \) = -P
=> 4 = -P
Product of roots \( \Large (2+i\sqrt{3}) \) \( \Large (2-i\sqrt{3}) \) = q,
=> 7 = -q.
Hence (p,q) = (-4, 7)