Can u find this relation?

The two roots satisfy:

$x = \frac{-p \pm \sqrt(p^2 - 4q)}{2}$ ;

and if one root is the square of the other, then we obtain:

$\frac{-p + \sqrt(p^2 - 4q)}{2} = (\frac{-p - \sqrt(p^2 - 4q)}{2})^2 \Rightarrow \frac{-p + \sqrt(p^2 - 4q)}{2} = \frac{p^2 + 2p*\sqrt(p^2 - 4q) + (p^2 - 4q)}{4}$ ;

or $\frac{-p + \sqrt(p^2 - 4q)}{2} = \frac{2p^2 + 2p*\sqrt(p^2 - 4q) - 4q)}{4}$ ;

or $2*[-p + \sqrt(p^2 - 4q)] = 2p^2 + 2p*\sqrt(p^2 - 4q) - 4q$ ;

or $(2 - 2p)\sqrt(p^2 - 4q) = 2p^2 + 2p - 4q$ ;

or $4(1 - 2p + p^2)*(p^2 - 4q) = (2p^2 + 2p - 4q)^2$ ;

or $4(p^2 - 2p^3 + p^4 - 4q + 8pq - 4p^2q) = 4p^4 + 8p^3 + 4p^2 - 16p^2q - 16pq + 16q^2$ ;

or $16p^3 - 48pq + 16q^2 + 16q = 0$ ;

or $p^3 - 3pq + q^2 + q = 0$ ;

or $\boxed{p^3 - q(3p - 1) + q^2 = 0}$ .