$$p + q = - p$$ and $$pq = q \Rightarrow q\left( {p - 1} \right) = 0$$

$$ \Rightarrow q = 0$$ or $$p=1.$$

If $$q = 0,$$ then $$p=0.$$ i.e.$$p=q$$

$$\therefore$$ $$p=1$$ and $$q=-2.$$

Let $$\alpha ,\beta $$ and $$\gamma ,\delta $$ be the roots of the equations $${x^2} + ax + b = 0$$

and $${x^2} + bx + a = 0$$ respectively.

$$\therefore$$ $$\alpha + \beta = - a,\alpha \beta = b$$

and $$\gamma + \delta = - b,\gamma \delta = a.$$

Given $$\left| {\alpha - \beta } \right| = \left| {\gamma - \delta } \right|$$

$$ \Rightarrow {\left( {\alpha - \beta } \right)^2} = {\left( {\gamma - \delta } \right)^2}$$

$$ \Rightarrow {\left( {\alpha + \beta } \right)^2} - 4\alpha \beta = {\left( {\gamma + \delta } \right)^2} - 4\gamma \delta $$

$$ \Rightarrow {a^2} - 4b = {b^2} - 4a$$

$$ \Rightarrow \left( {{a^2} - {b^2}} \right) + 4\left( {a - b} \right) = 0$$

$$ \Rightarrow a + b + 4 = 0$$

( as $$a \ne b$$ )

Product of real roots $$ = {9 \over {{t^2}}} > 0,\forall \,t\, \in R$$

$$\therefore$$ Product of real roots is always positive.

We have $${\alpha ^2} = 5\alpha - 3$$ and $${\beta ^2} = 5\beta - 3;$$

$$ \Rightarrow \alpha \,\,\& \,\,\beta $$ are roots of

equation, $${x^2} = 5x - 3$$ or $${x^2} - 5x + 3 = 0$$

$$\therefore$$ $$\alpha + \beta = 5$$ and $$\alpha \beta = 3$$

Thus, the equation having $${\alpha \over \beta }\,\,\& \,\,{\beta \over \alpha }$$ as its roots is

$${x^2} - x\left( {{\alpha \over \beta } + {\beta \over \alpha }} \right) + {{\alpha \beta } \over {\alpha \beta }} = 0$$

$$ \Rightarrow {x^2} - x\left( {{{{\alpha ^2} + {\beta ^2}} \over {\alpha \beta }}} \right) + 1 = 0$$

or $$3{x^2} - 19x + 3 = 0$$