Your intuition betrays you.
Under any reasonable computational model of real numbers, the following are undecidable:
- Is a real number equal to zero?
- Is a real number positive?
- Is a $2 \times 2$ matrix invertible?
- Does a quadratic equation with real coefficients have two distinct roots?
All of the above have the computational strength of a Halting oracle. For example, here is how one can reduce the Halting problem to zero-testing. Given a Turing machine $T$, define the following sequence of rationals:
$$
q_n = \begin{cases}
2^{-n} & \text{if $T$ has not halted by step $n$,} \\
2^{-k} & \text{if $T$ halts in step $k \leq n$.}
\end{cases}
$$
In words, $q_0, q_1, q_2, \ldots$ falls off to $0$ for as long as $T$ is still running, and if and when $T$ halts, the sequence stabilizes at a positive value.
Because $q_n$ is a computable Cauchy sequence with a computable modulus of convergence, we can compute its limit $x = \lim_n q_n$. Now we have $$x \neq 0 \iff \text{$T$ halts}.$$
Let us talk about the definition of "reasonable":
The usual structure of the field of reals is computable: arithmetical operations $+$, $\times$, $0$, $1$, $-$, ${}^{-1}$.
The order-theoretic structure is computable: $\min$, $\max$, and the strict $<$ is semi-decidable.
An essential characteristic of the reals is that they are complete. A reasonable model of real-number computation therefore provides such a notion. It can be Dedekind completeness (every double-sided real cut determines a unique real), Cauchy completeness (every Cauchy sequence has a limit), or if one is careful enough even MacNielle completeness (an inhabited bounded set has a supremum). However, such completeness must be realized in a computable way. What precisely that means depends on the computational model.
The computational model must be realistic in the sense that it does not enable computations which are not computable in the sense of Turing.
Note that the above conditions do not imply that every real is representable. There are reasonable models in which only the computable reals are accounted for, but also ones that can represent all reals. In every case, the resulting structure is (computably) a complete ordered Archimedean field.
Here are some unreasonable models:
Violating computability of basic arithmetic: the usual decimal expansion (because $+$ is not computable).
Violating completeness: real closed field, algebraic numbers, fixed-precision floating points.
Violating Turing computability: the so-called real RAM model, also known as Blum-Shub-Smale, because it has builtin decidability of $<$ (and therefore also of $=$ and $\leq$).