Complex Numbers And Quadratic Equations question 543
Question: If the roots of the equation $ \frac{1}{x+p}+\frac{1}{x+q}=\frac{1}{r} $ are equal in magnitude but opposite in sign, then the product of the roots will be [IIT 1967; RPET 1999]
Options:
A) $ \frac{p^{2}+q^{2}}{2} $
B) - $ \frac{(p^{2}+q^{2})}{2} $
C) $ \frac{p^{2}-q^{2}}{2} $
D) - $ \frac{(p^{2}-q^{2})}{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
Given equation can be written as $ x^{2}+x(p+q-2r)+pq-pr-qr=0 $ …..(i) whose roots are $ \alpha $ and $ -\alpha $ , then the product of roots $ -{{\alpha }^{2}}=pq-pr-qr=pq-r(p+q) $ …..(ii) and sum $ 0=p+q-2r\Rightarrow r=\frac{p+q}{2} $ …..(iii) From (ii) and (iii), we get $ -{{\alpha }^{2}}=pq-\frac{p+q}{2}(p+q)=-\frac{1}{2}{ {{(p+q)}^{2}}-2pq } $ $ =-\frac{(P^{2}+q^{2})}{2} $ .
BETA
