Let $F$ be a field of characteristic not 2, and let $D_1$ and $D_2$ be non-squares in $F$. Prove that $F(\sqrt{D_1}, \sqrt{D_2})$ has degree 4 over $F$ if $D_1D_1$ is nonsquare in $F$, and degree 2 otherwise. (If the degree is 4, $F(\sqrt{D_1},\sqrt{D_2})$ is called a biquadratic extension of $F$.
Suppose $D_1D_2$ is nonsquare in $F$. We claim that $p(x) = x^2 - D_2$ is irreducible over $F(\sqrt{D_1})$. To see this, suppose to the contrary that $(a+b\sqrt{D_1})^2 = D_2$. Comparing coefficients, we have $a^2 + b^2D_1 - D_2 = 0$ and $2ab = 0$. If $b = 0$, then $D_2 = a^2$, a contradiction. If $a = 0$, then $D_1 = D_2/b^2$. But then $D_1D_2 = (D_2/b)^2$ is square, a contradiction. So $F(\sqrt{D_1},\sqrt{D_2})$ has degree 2 over $F(\sqrt{D_1})$, and thus degree 4 over $F$.
Now suppose that $D_1D_2 = \eta^2$ is square in $F$. then $(x - \eta/\sqrt{D_1})(x + \eta/\sqrt{D_1}) = x^2 - D_2$ is reducible over $F(\sqrt{D_1})$, and we have $F(\sqrt{D_1},\sqrt{D_2}) = F(\sqrt{D_1})$ of degree 2 over $F$.