If 2+i√3 is a root of the equation x2+px+q=0, where p and q are real, then (p, q) =
(-4,-7)
(4,7)
(-4,7)
Since 2+i√3 is a root, therefore 2−i√3 will be other root. Now sum of the roots = 4 = -p and product of roots = 7 = q. Hence (p, q) = (-4, 7).
If (2+i√3) is a root of the equation x2+px+q=0 where p and q are real, then find (p,q).