If roots of \(\frac{1}{x+p}+\frac{1}{x+q}=\frac{1}{r}\) are equal in magnitude and opposite in sign, then which of the following is true ?

Given:

The roots of \(\frac{1}{x+p}+\frac{1}{x+q}=\frac{1}{r}\) are equal in magnitude and opposite in sign

Concept:

For any quadratic equation let us say

ax² + bx + c = 0

Sum of the roots = -b/a

Product of the root = c/a

Solution:

\(\frac{1}{x+p}+\frac{1}{x+q}=\frac{1}{r}\)

⇒ \(2x+p+q\:=\:\frac{(x+p)(x+q)}{r}\)

⇒ \(r(2x+p+q)\:=\:x^2+(p+q)x+pq\)

⇒ \(x^2+(p+q-2r)x+[pq+(p+q)r]=0\)

Comparing above with ax2 + bx + c = 0 we will get, 

a = 1, b = (p + q -2r), c = [pq + (p + q) r]

According to the question, the roots of the given equation are equal in magnitude and opposite in sign

Let say \(\alpha \:and \: -\alpha\) be the roots, then

Sum of the roots = [\(\alpha \:+ \: (-\alpha)\)] = 0 = -b/a

⇒ -b/a = -(p + q - 2r) = 0

⇒ p + q = 2r