Note: this question was updated (2) after GNiklasch's answer was posted, and taking Gro-Tsen's comment into account. The initial question (1) dealt with $\mathbb{Q}_3$.

**Original post (1).**Let's try to solve the equation $x^2 - 2 = 0$ with $x = \frac{a}{b} \in \mathbb{Q}$. We can't have $x^2 \neq 2$, so the best we can do is minimize $|x^2 - 2|$. Let's try to find an approximation that works over two different completions. Can we have this?

$$ |(\tfrac{a}{b})^2 - 2 |_\infty \ll \frac{1}{a} \text{ and } |(\tfrac{a}{b})^2 - 2 |_3 \ll \frac{1}{a} $$

I'm trying to write an $S$-adic approximate solution over two places $S = \{ 3, \infty\}$ and $x = \frac{a}{b} \mapsto (\frac{a}{b}, \frac{a}{b}) \in \mathbb{Q} \times \mathbb{Q} \subset \mathbb{R} \times \mathbb{Q}_3 $. What are the correct exponents?

**Edit (2).** Perhaps i need to find a problem statement that has a solution. gro-tsen suggests I change $p=3$ to $p=7$ so that $\sqrt{2}\in \mathbb{Q}_7$.

$$ |(\tfrac{a}{b})^2 - 2 |_\infty \ll \frac{1}{a} \text{ and } |(\tfrac{a}{b})^2 - 2 |_7 \ll \frac{1}{a} $$

possibly I can leave the places the same and change the thing I'm approximating. $\sqrt{7}\in \mathbb{Q}_3$ so perhaps I can find a rational number $\frac{a}{b}\in \mathbb{Q}$ such that

$$ |(\tfrac{a}{b})^2 - 2 |_\infty \ll \frac{1}{a} \text{ and } |(\tfrac{a}{b})^2 - 7 |_3 \ll \frac{1}{a} $$

Excuse me while I try to state an instance of weak approximation that's not vacuous.