2. Determine the largest integer $M>1$ with the following property: for any 10 different real numbers chosen from the interval $[1, M]$, it is always possible to select three numbers, denoted as $a<b<c$, such that the quadratic equation $a x^{2}+b x+c=0$ has no real roots.
First, prove that $M=4^{255}$ has the property described in the problem, using proof by contradiction.
Assume $a_{1}4^{2^{9-k}-1}$. When $k=9$, since $a_{10}>a_{9} \Rightarrow \frac{a_{10}}{a_{9}}>1$, the conclusion holds;
Assume for some $2 \leqslant k \leqslant 9$, $\frac{a_{10}}{a_{k}}>4^{2^{9-k}-1}$, then for $k-1$, construct the equation $a_{k-1} x^{2}+a_{k} x+a_{10}=0 \Rightarrow a_{k}^{2}-4 a_{k-1} a_{10} \geqslant 0 \Rightarrow \frac{a_{k}}{a_{k-1}} \geqslant 4 \cdot \frac{a_{10}}{a_{k}}>4^{2^{9-k}}$, thus $\frac{a_{10}}{a_{k-1}}=\frac{a_{10}}{a_{k}} \cdot \frac{a_{k}}{a_{k-1}}>4^{2^{9-k}-1+2^{9-k}}=4^{2^{10-k}-1}$, i.e., the conclusion also holds for $k-1$. By the principle of mathematical induction, for $1 \leqslant k \leqslant 9$, $\frac{a_{10}}{a_{k}}>4^{2^{9-k}-1}$. In particular, when $k=1$, $\frac{a_{10}}{a_{1}}>4^{2^{8}-1}=4^{255}$, which contradicts $a_{1}, a_{10} \in[1, M] \Rightarrow \frac{a_{10}}{a_{1}} \leqslant M=4^{255}$.
Second, when $M>4^{255}$, let $M=4^{255} \lambda^{256}, \lambda>1$, and choose $a_{1}, a_{2}, \cdots, a_{10}$ as follows: $a_{10}=M, a_{k}=\frac{a_{10}}{4^{2^{9-k}-1} \lambda^{2^{9-k}}}, 1 \leqslant k \leqslant 9$.
Clearly, $1=a_{1}4^{255}$ does not have the property described in the problem. In summary, the largest integer $M=4^{255}$.