Problem: If a±bi(bî€ =0) are imaginary roots of the equation x3+qx+r=0 where a,b,q, and r are real numbers, then q in terms of a and b is
Answer Choices:
A. a2+b2
B. 2a2−b2
C. b2−a2
D. b2−2a2
E. b2−3a2
Solution:
Let the third rook of the given equation be s . The sum of the roots is zero, 2a+s=0,s=−2a, The sum of the producte of the roots taken two at a time is q.
(a+bi)(a−bi)+(a+bi)s+(a−bi)s=q
(a2+b2)+a(−2a)+bi(−2a)+a(−2a)−bi(−2a)=b2−3a2=q
which is choice (E).