Answer: \(p^2+4r=0\), and \(q=0\).

Let the two pairs of equal roots be \(\alpha,\alpha,\beta,\beta\).

For \(x^4-px^2+qx-r=0\), Vieta's formulae give

\(2\alpha+2\beta=0\), so \(\alpha+\beta=0\), hence \(\beta=-\alpha\).

The sum of the pairwise products is

\(\alpha^2+4\alpha\beta+\beta^2=-p\).

Using \(\beta=-\alpha\),

\(-p=\alpha^2-4\alpha^2+\alpha^2=-2\alpha^2\), so \(p=2\alpha^2\).

The product of the roots is

\(\alpha^2\beta^2=-r\).

Again using \(\beta=-\alpha\), this gives \(\alpha^4=-r\).

Therefore

\(p^2=(2\alpha^2)^2=4\alpha^4=-4r\),

so \(p^2+4r=0\).

For \(q\), the sum of the triple products is

\(2\alpha^2\beta+2\alpha\beta^2=-q\).

This is \(2\alpha\beta(\alpha+\beta)\), which is zero because \(\alpha+\beta=0\). Hence \(q=0\).