Complex Numbers And Quadratic Equations question 755
Question: The solution of equation $ \frac{p+q-x}{r}+\frac{q+r-x}{p} $ + $ \frac{r+p-x}{q} $ + $ \frac{4x}{p+q+r}=0 $ is [MP PET 2004]
Options:
A) $ x=p+q+r $
B) $ x=p-q+r $
C) $ x=\frac{p+q}{q+r} $
D) $ x=\frac{p}{q}+r $
Show Answer
Answer:
Correct Answer: A
Solution:
We have $ \frac{p+q-x}{r}+\frac{q+r-x}{p}+\frac{r+p-x}{q} $ = $ \frac{-4x}{p+q+r} $ $ \frac{p+q+r-x}{r}+\frac{p+q+r-x}{p}+\frac{p+q+r-x}{q} $ = $ 4-\frac{4x}{p+q+r} $
$ \Rightarrow $ $ (p+q+r-x)( \frac{1}{p}+\frac{1}{q}+\frac{1}{r} )=4( \frac{p+q+r-x}{p+q+r} ) $
$ \Rightarrow $ $ (p+q+r-x)[ \frac{1}{p}+\frac{1}{q}+\frac{1}{r}-\frac{4}{p+q+r} ]=0 $
$ \Rightarrow $ $ x=p+q+r $ .
BETA
