 Solution

Let $f(x)=a{\prod }_{i=1}^n(x-{\alpha }_i)$. Its polynomial discriminant isThe squared Vandermonde determinant is a symmetric polynomial in the roots. By the Fundamental theorem of symmetric polynomials, it is a polynomial in their elementary symmetric functions, which are the coefficients of $f$ by Vieta formulas; hence $\Delta (f)\in K$. Equivalently, the resultant formula puts it directly in $K$.

For $f=x^4+rx+s$, a resultant calculation with $f^{\prime }=4x^3+r$ gives