Number of points on an elliptic curve and it's twist over $\mathbb{F}_p$. I have another probably very trivial question about elliptic curves. This wikipedia article gives the following formula $|E|+|E^d|=2p+2$ where $E$ is an elliptic curve over $\mathbb{F}_p$ and $E^d$ is it's quadratic twist. At the same time the paragraph preceding this statement seems to argue why this should be true making it seem easy, but I seem to be missing a step or two. I'll try to reconstruct the proof and finish up by asking about the step I'm unable to get. In all that follows I assume $p$ is a prime greater then $3$ and all my elliptic curves are over $\mathbb{F}_p$.
Let us have $E:y^2=x^3+ax+b$ and it's elliptic twist $E^d:dy^2=x^3+ax+b$ where $d$ is a non-square in $\mathbb{F}_p$. Then for every $x\in\mathbb{F}_p$ we have $x^3+ax+b$ or $d^{-1}(x^3+ax+b)$ is a square and so there is a $y$ such that $(x,y)\in E$ or $(x,y)\in E^d$. Furthermore if $x^3+ax+b\neq 0$ then there exist exactly two such $y$ (namely $\pm\sqrt{x^3+ab+b}$ since if $y=-y$ we get $y+y=0$ and since $p>3$ we get $y=0$).
This gives us almost the result we want except we might have up to 3 $x$'s where $x^3+ax+b=0$ which only have 1 solution. How do we fix this to get us the wanted result?

If $x^3+ax+b = 0$, then the point $(x,0)$ lies on each of $E$ and $E^d$, for a total of $2$ solutions.
