### 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