Complex Numbers And Quadratic Equations question 127
Question: Given that the equation $ z^{2}+(p+iq)z+r+is=0, $ where $ p,q,r,s $ are real and non-zero has a real root, then
Options:
A) $ pqr=r^{2}+p^{2}s $
B) $ prs=q^{2}+r^{2}p $
C) $ qrs=p^{2}+s^{2}q $
D) $ pqs=s^{2}+q^{2}r $
Show Answer
Answer:
Correct Answer: D
Solution:
Given that $ z^{2}+(p+iq)z+r+is=0 $ ……(i) Let $ z=\alpha $ (where $ \alpha $ is real) be a root of (i), then $ {{\alpha }^{2}}+(p+iq)\alpha +r+ $ is =0 or $ {{\alpha }^{2}}+p\alpha +r+i(q\alpha +s) $ =0 Equating real and imaginary parts, we have $ {{\alpha }^{2}}+p\alpha +r=0 $ and $ q\alpha +s=0 $ Eliminating $ \alpha , $ we get $ {{( \frac{-s}{q} )}^{2}}+p( \frac{-s}{q} )+r=0 $ or $ s^{2}-pqs+q^{2}r=0 $ or $ pqs=s^{2}+q^{2}r $
BETA
