If I was trying to construct a splitting field for $(x^2-3)(x^2-5)$ over $\Bbb Q(\sqrt{2})$.
Then obviuously I would begin by checking if $\sqrt3$ was an element of $\Bbb Q(\sqrt{2})$.
To do this I would assume it is get the equation
$$3=a^2+2ab+b^2$$
and show that this is impossible for the three cases $a=0,b=0,ab=0$
And so adjoin this element to get
$$\Bbb Q(\sqrt{3},\sqrt{2}):=\{a+b\sqrt2+c\sqrt3+\sqrt{6}|a,b,c,d \in \Bbb Q\}$$
But proving that $\sqrt{5}\notin \Bbb Q(\sqrt{3},\sqrt{2})$ in the same manner as above , although rather straightforward, is messy and time consuming. Therefore in an exam setting I would like to avoid this method.
Is there a more elegant method to show that $\sqrt{5}\notin \Bbb Q(\sqrt{3},\sqrt{2})$ ?
Seems like you do not have a complete toolkit yet, and I guess proving from the definition is probably the only hope. But let me try to make it look less painful.
If $\sqrt 5 \in \mathbb Q[1, \sqrt2, \sqrt3, \sqrt6]$, so is $\sqrt{\alpha}$ for $\alpha \in \{5,10,15,30\}$. Thus, assume
$$ \sqrt\alpha = a + b\sqrt2 + c\sqrt3 + d\sqrt 6, $$
where $a,b,c,d \in \mathbb Q$ and $a \neq 0$. Since $\sqrt\alpha^2 \in \mathbb Q$, by comparing terms we have
$$ \begin{aligned} 2bc + 2ad &= 0;\\ 2ab + 6cd &= 0;\\ 2ac + 4bd &= 0. \end{aligned} $$
From the first equation $d = -\frac{bc}{a}$. Plug into the second, $b(a^2 - 3c^2)=0$. Since $a \neq 0$, $b = 0$, so $d =0$, so $c = 0$, so $\sqrt \alpha \in \mathbb Q$, contradiction.