This question originates from Pinter's Abstract Algebra, Chapter 27, Exercise F3.
Let $F$ be a finite field. If $a, b \in F$, let $p(x) = x^2-a$ and $q(x) = x^2 - b$ be irreducible in $F[x]$, and let $\sqrt{a}$ and $\sqrt{b}$ denote roots of $p(x)$ and $q(x)$ in an extension of $F$.
Exercise F2 shows why $a/b$ is a square, say $a/b = c^2$ for some $c \in F$; and that $\sqrt{b}$ is a root of $p(cx)$.
Prove that $F[x]/\langle p(cx)\rangle\cong F(\sqrt{b})$, and $F(\sqrt{a}) \cong F(\sqrt{b})$.
By Exercise F2, $\sqrt{b}$ is a root of $p(cx)$, which implies $F[x]/\langle p(cx)\rangle\cong F(\sqrt{b})$.
By Exercise E5, $F(\sqrt{a}) \cong F[x]/\langle p(x)\rangle = F[x]/\langle p(cx)\rangle \cong F(\sqrt{b})$.
Correct?