JEE Main On 16 April 2018 Question 6
Question: Let p, q, and r be real numbers $ (p\ne q,r\ne 0), $ such that 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 sum of squares of these roots is equal to. [JEE Main 16-4-2018]
Options:
A) $ p^{2}+q^{2}+r^{2} $
B) $ p^{2}+q^{2} $
C) $ 2(p^{2}+q^{2}) $
D) $ \frac{p^{2}+q^{2}}{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \frac{1}{x+p}+\frac{1}{x+q}=\frac{1}{r} $
$ \frac{x+p+x+q}{(x+p)(x+q)}=\frac{1}{r} $
$ (2x+p+q)r=x^{2}+px+qx+pq $
$ x^{2}+(p+q-2r)x+pq-pr-qr=0 $ Let and be the roots.
$ \Rightarrow $ $ \alpha +\beta =-(p+q-2r) $ …[1]
$ \Rightarrow $ $ \alpha \beta =pq-pr-qr $ …[2] Roots are equal in magnitude and opposite in sign
$ \Rightarrow $ $ \alpha +\beta =0. $
$ \Rightarrow $ $ -(p+q-2r)=0 $ …[3]
$ {{\alpha }^{2}}{{\beta }^{2}}={{(\alpha +\beta )}^{2}}-2\alpha \beta $
$ ={{(-(p+q-2r))}^{2}}-2(pq-pr-qr) $ …(from [1] and [2])
$ =p^{2}+q^{2}+4r^{2}+2pq-4pr-4qr-2pq+2pr+2qr $ $ =p^{2}+q^{2}+2r(2r-p-q) $ …(from[3])
$ =p^{2}+q^{2}+0 $ $ =p^{2}+q^{2} $
Hence, answer is option B
