I'm reading Schütt-Shioda's survey paper on elliptic surfaces, section "Tate's algorithm".
My question doesn't concern the algorithm itself, but only the initial manipulation the authors do on the Weierstrass equation.
If $f:S\to C$ is an elliptic fibration, the generic fiber and $\Delta$ are given by: $$y^2=x^3+a_2x^2+a_4x+a_6$$ $$\Delta=-27a_6^2+18a_2a_4a_6+a_2^2a_4^2-4a_2^3a_6-4a_4^3$$
We want to analyze when $\Delta=0$. We take a local parameter $t$ in $C$ and a valuation $v$ with $v(t)=1$, so $\Delta(0)=0$ is equivalent to $v(\Delta)>0$.
The author then says: "by a translation in $x$, we can move the singularity to $(0,0)$. Then the extended Weierstrass form transforms as $y^2=x^3+a_2x^2+ta'_4x+ta_6'$".
I don't understand how the translation is done. If we translate $x$ by $\lambda$, we get a new Weierstrass equation with free term $\lambda^3+a_2\lambda^2+a_4\lambda+a_6$, which is zero iff $\lambda$ is a root of $T^3+a_2T^2+a_4T+a_6$. But the existence of such $\lambda$ is highly dependent on the field, right? Besides, assuming this is possible, I do understand why we change $a_6$ for $ta_6'$, but not why we change $a_4$ for $ta_4'$.