Let $a,b,c,d \ge 1$ be integers with $b$ and $d$ nonsquare and $a\sqrt{b} \ge c\sqrt{d}$.
Now I have three related questions:
Under what conditions can one find $u,v,w$ such that $a\sqrt{b} \pm c\sqrt{d} = u+v\sqrt{w}$?
In particular, under what circumstances can $u=0$?
If such a representation exists, what would be considered the "minimal" solution $(u,v,w)$?
A slightly more general version of your problem would be "For arbitrary integers $a_i$ and distinct nonzero squarefree integers $n_i$ (1 is still allowed), when can $a_1\sqrt{n_1} + a_2\sqrt{n_2} + \dots + a_k \sqrt {n_k}$ be equal to 0? The answer is when all $a_i = 0$.
Let $P = p_1, p_2, \dots$ be the list of prime factors of $n_1$ through $n_k$, in arbitrary order.
Let $S_0 = \mathbb{Z}$. Let $S_1$ be the arithmetic closure of $S_0 \cup \{ \sqrt{p_1} \}$ over $+$, $-$, and $\cdot$. Let $S_2$ be the arithmetic closure of $S_1 \cup \{ \sqrt{p_2} \}$, and so on.
Since $S_0$ is $\mathbb{Z}$, it obviously only has one representation for a zero element, the integer 0.
Every element of $S_1$ can be represented as $a + b \sqrt{p_1}$ where $a, b \in S_0$. The interesting question here is under which circumstances does $a + b \sqrt{p_i} = 0$?
Therer are nine cases to consider: $\{ a < 0, a = 0, a > 0 \} \times \{ b < 0, b = 0, b > 0\}$. Obviously, for $(a = 0, b = 0)$ the quantity is also zero. None of the other cases are interesting except $(a < 0, b > 0)$ and $(a > 0, b < 0)$.
Given $a < 0$ and $b > 0$, it's obvious that $-a + b\sqrt{p_i} > 0$. So $$\begin{align} a + b\sqrt{p_i} = 0 & \Leftrightarrow (a + b\sqrt{p_1})(-a + b\sqrt{p_1}) = 0\\ & \Leftrightarrow b^2p_1 - a^2 = 0 \end{align} $$
Since $b^2p_1 - a^2$ can only be 0 when $b=0$ and $a=0$, and we know this contradicts our case assumptions, we know that $a + b\sqrt{p_i}$ can't be zero when $a < 0$ and $b > 0$. Likewise, it can't be zero when $a > 0$ and $b < 0$.
This proof can be continued inductively. Every element in $S_2$ can be represented in the form $a + b\sqrt{p_2}$ where $a, b \in S_1$. $S_1$ has a unique zero element. $a + b\sqrt{p_2}$ can only be 0 when both $a=0$ and $b=0$, so $S_2$ has a unique zero element.