But if the given curve has a node, then not only the Hessian passes through the node, but it has there a node the two branches at which touch respectively the two branches of the curve; and the node thus counts as six intersections; so if the curve has a cusp, then the Hessian not only passes through the cusp, but it has there a cusp through which it again passes, that is, there is a cuspidal branch touching the cuspidal branch of the curve, and besides a simple branch passing through the cusp, and hence the cusp counts as eight intersections.
This method of solution fails when the discriminant R vanishes, for then the Hessian has equal roots, as also the cubic f.
The Hessian in that case is a factor of f, and Q is the third power of u2,...
The quartic has four equal roots, that is to say, is a perfect fourth power, when the Hessian vanishes identically; and conversely.
This can be verified by equating to zero the five coefficients of the Hessian (ab) 2 axb2.