Complex Numbers And Quadratic Equations question 820
Question: The solution set of the equation $ pqx^{2}-{{(p+q)}^{2}}x+{{(p+q)}^{2}}=0 $ is [Kerala (Engg.) 2005]
Options:
A) $ { \frac{p}{q},\frac{q}{p} } $
B) $ { pq,\frac{p}{q} } $
C) $ { \frac{q}{p},pq } $
D) $ { \frac{p+q}{p},\frac{p+q}{q} } $
E) $ { \frac{p-q}{p},\frac{p-q}{q} } $
Show Answer
Answer:
Correct Answer: D
Solution:
Given equation $ (pq)x^{2}-{{(p+q)}^{2}}x+{{(p+q)}^{2}}=0 $ Let solution set is $ { \frac{p+q}{p},\frac{p+q}{q} } $ Sum of roots = $ \frac{{{(p+q)}^{2}}}{pq} $
Þ $ \frac{p+q}{p}+\frac{p+q}{q}=\frac{{{(p+q)}^{2}}}{pq} $ Similarly, product of roots = $ \frac{{{(p+q)}^{2}}}{pq} $
Þ $ \frac{p+q}{p}\times \frac{p+q}{q}=\frac{{{(p+q)}^{2}}}{pq} $ .
BETA
