Solution set of $\dfrac{a}{x} + \dfrac{a}{x + 1} = \dfrac{2a}{x - 1}$ leads to a quadratic in $x$ whose constant term si

Question

Solution set of $\dfrac{a}{x} + \dfrac{a}{x + 1} = \dfrac{2a}{x - 1}$ leads to a quadratic in $x$ whose constant term sign is—

Accepted Answer

negative

Answer

positive

Answer

zero

Answer

depends on $a$

Solution set of ax+ax+1=2ax−1\dfrac{a}{x} + \dfrac{a}{x + 1} = \dfrac{2a}{x - 1}xa​+x+1a​=x−12a​ leads to a quadratic in xxx whose constant term sign is—