For a binary quadratic form $ax^2+bxy+cy^2$ over a field (characteristic not 2), the discriminant $b^2-4ac$ is 0 if and only if that form is the square of a linear form. I am curious about an analogue of this fact over a ring with a valuation.

* Rough version of question. * Let
$$f=ax^2+bxy+cy^2$$
be a binary quadratic form over a ring $Q$ with a valuation. If the valuation of the discriminant $b^2-4ac$ is large (so heuristically $b^2-4ac$ is ``almost 0''), does this imply that $f$ is "almost a square"?

* More precise version of question.* Here's the precise case I am interested in. Let $R$ be a valuation ring, where 2 is a unit. Let $Q$ be the standard graded ring $Q=R[s,t]$ over $R$, i.e. $\text{deg}(s)=\text{deg}(t)=1$ and $Q_0=R$. The ring $Q$ inherits a valuation from $R$. Let $f$ as above, where $a$,$b$, and $c$ are all homogeneous elements of $Q$, and where $\text{deg}(a)+\text{deg}(c)=2\cdot \text{deg}(b).$

** Claim: ** If $b^2-4ac$ has valuation at least $\nu$, then there exist $d,e,f$, and $a',b',c'$ in $Q$, such that
$$
ax^2+bxy+cy^2=d(ex+fy)^2 + (a'x^2+b'xy+c'y^2),
$$
where $a',b',$ and $c'$ all have valuation at least $\nu/2$.

** Update: ** Let me add some clarification and an example. First of all, note that $Q$ is not a valuation ring, since $Q=R[s,t]$. So an element with valuation $0$ is not necessarily a unit. For instance, $s^2$ is an element of $Q$ with valuation $0$ that is not a unit.

Also, here's an example. Let $R=\mathbb Q[[u]]$ and let $f=u^3x^2+u^3y^2$. Then $b^2-4ac=-4u^6$, which has valuation $6$. We can (trivially) write $f=(0)^2+f$ as the sum of a square and something with valuation $6/2=3$. So I think that the $\nu/2$ bound in the claim is optimal.