Know the logic?

Given that the roots are $\dfrac{p}{q}$ and $-\dfrac{q}{p}$ , we know that:

$\left( x - \dfrac{p}{q} \right) \left( x + \dfrac{q}{p} \right) = 0\\ x^2 - \dfrac{p}{q}x + \dfrac{q}{p}x - 1 = 0\\ x^2 - \left (\dfrac{p^2}{pq} - \dfrac{q^2}{pq} \right) x - 1 = 0\\ \boxed{pqx^2 - (p^2 - q^2)x - pq = 0}$

Whenever the roots are reciprocal the each other then $LaTeX$ c=a. $LaTeX$ ( product of the roots is $LaTeX$ c/a $LaTeX$ ..) So here roots are reciprocal to each other and are opposite in sign.. so $LaTeX$ a= -c $LaTeX$ . So this is only satisfied by $LaTeX$ pqx^2- ( p ^2-q^2)x- pq $LaTeX$

Lets see what the sum of the roots are :- $\dfrac{p}{q} - \dfrac{q}{p} = \dfrac{p^2 - q^2}{pq}$
Now, we know that for a quadratic equation $ax^2 + bx + c$ , the sum of the roots = $-\dfrac{b}{a}$
Trial and error application of this formula in the given options show that only one options satisfies the condition that the sum of roots is $\dfrac{p^2 - q^2}{pq}$ and that is $pqx^2 - (p^2 - q^2)x - pq$ .

How to directly find which of the given options satisfy this value?
P.S. This trick might not work in some cases.
The denominator of the sum should have $pq$ and only 2 such options have the values and those options have the magnitude of the coefficient of x raised to the power 1 as $(p^2 - q^2)$ . And as the formula of the sum is $-\dfrac{b}{a}$ , the correct option should have a negative sign befire the coefficient of the variable with degree 1 i.e. coefficient of x. And only one such option have that value and so the answer is $\boxed{pqx^2 - (p^2 - q^2)x - pq}$ .