Problem: If aΒ±bi(bξ =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. 2a2βb2
C. b2βa2
D. b2β2a2
E. b2β3a2
Solution:
Let the third rook of the given equation be s . The sum of the roots is zero, 2a+s=0,s=β2a, The sum of the producte of the roots taken two at a time is q.
(a+bi)(aβbi)+(a+bi)s+(aβbi)s=q
(a2+b2)+a(β2a)+bi(β2a)+a(β2a)βbi(β2a)=b2β3a2=q
which is choice (E).