Let $disc(f) = D = \Delta^2$ be denoted as the discriminant of $f$ and assume that $\Delta \notin K.$ Let $g \in K[x]$ be the resolvent cubic of $f$ and assume there exists only one root which is $d \in K$
Questions:
a) If an element $p \in K$ is not a square in $K$ but is a square in $K(\Delta)$, then show that $p = r^2\Delta^2$ for some $r \in K$
b) Let $x^2 + d$ and $x^2 - (a - d)x + c$ be polynomials that split over $K(\Delta)$: show that the discriminant of the two is either $0$ or not a square in $K$
For part a) I know that: $\Delta = \prod_{1 \leq i \leq j \leq n} (x_j - x_i)$
So by definition there exists a map $\sigma$ which maps $\Delta$ to $ \pm \Delta.$
Now I'm stuck here because $p$ is not a square in $K$ so I'm not sure what to do.
For part b) I directly tried to compute for the discriminant. So $x^2 + d$ always has repeated roots so the discriminant is $0$ for that. But I'm unsure how to approach $x^2 - (a - d)x + c$, any suggestions?