Transforming Quadrics in Characteristic 2 I’m trying to solve the following problem given in a textbook:

Let $k$ be an algebraically closed field and $Q=V(F)$ a quadric in $\mathbb{P}^3(k)$, where $F$ is an irreducible polynomial in $X,Y,Z,T$, and hence gives rise to a quadratic form on $k^4$ which we assume is non-degenerate.
Show that after some change of coordinates we can write $Q=V(XT-YZ)$.

I’ve solved the case where $\text{char}(k)\neq2$ by diagonalising the quadratic form and then making a suitable change of coordinates. However this process involves using the correspondence between quadratic forms and symmetric bilinear forms which is not valid in characteristic $2$ (and if we could diagonalise $F$ then it would become reducible).
My first issue is that I can’t find a reference defining what it means for a quadratic form to be non-degenerate in characteristic $2$, so if I tried to come up with a counterexample I wouldn’t know if it was valid or not.
Beyond this, I’m not even sure that the result is true, I can’t seem to find anywhere claiming that it is. Has the textbook simply forgotten to specify that $\text{char}(k)\neq2$, or am I missing something?
Any help would be much appreciated.
 A: Having spent more time on this, I think I have solved the problem, and the result is in fact true.
Arf defines non-singularity for quadratic forms of characteristic $2$ here, explained in English here. From these papers, we see that if $Q$ is non-singular over a field of characteristic $2$, we can write $$Q=(aX^2+XT+bT^2)+(cY^2+YZ+dZ^2)$$ for some $a,b,c,d\in k$ and an appropriate choice of coordinates. Let’s consider $aX^2+XT+bT^2$.
If $a=b=0$ then we have $XT$ already, if say $b=0$ then we have $X(aX+T)$ so taking the inverse of the transformation sending $X\mapsto X$ and $T\mapsto aX+T$ we have $XT$.
Then we assume $a,b\neq0$. Sending $X\mapsto\frac{1}{\sqrt{a}}X$ and $T\mapsto\frac{1}{\sqrt{b}}T$ we have $X^2+\alpha XT+T^2$ for $\alpha=\frac{1}{\sqrt{ab}}$. All square roots exist since $k$ is algebraically closed, and for the same reason we can also find a root $\beta$ of the polynomial $x^2+\alpha x+1$. Then sending $X\mapsto X+\frac{1}{\alpha^2}T$ and $T\mapsto\beta X+\frac{\alpha+\beta}{\alpha^2}T$ we have $XT$. This transformation is invertible since $$\begin{vmatrix}1 & \beta \\\frac{1}{\alpha^2} & \frac{\alpha+\beta}{\alpha^2}\end{vmatrix}=\frac{1}{\alpha}\neq0$$
We can repeat the same process for $Y$ and $Z$, and so we can write $Q=XT+YZ=XT-YZ$.
