Nếu b < 0 thì $\underset{x\to +\infty }{\lim }\sqrt{c{x}^2+ax}-bx=+\infty \Rightarrow b>0$ khi đó ta có

$$\begin{gathered} \underset{x\to +\infty }{\lim }\sqrt{c{x}^2+ax}-bx=2\iff \underset{x\to +\infty }{\lim }\frac{c{x}^2+ax-b^2{x}^2}{\sqrt{c{x}^2+ax}+bx}=2 \\ \iff \underset{x\to +\infty }{\lim }\frac{x\left(cx+a-b^{2}x\right)}{x\left(\sqrt{c+\frac{a}{x}}+b\right)}=2\iff \underset{x\to +\infty }{\lim }\frac{cx+a-b^{2}x}{\sqrt{c+\frac{a}{x}}+b}=2 \\ \iff \left\{\begin{array}{l}c-b^2=0\\ \frac{a}{\sqrt{c}+b}=2\end{array}\iff \left\{\begin{array}{l}c-b^2=0\\ a=2(\sqrt{c}+b)\end{array}\right.\right. \end{gathered}$$

Kết hợp ${\text{b}}^2+\text{c}=8\Rightarrow \text{c}=4;\text{b}=2;\text{a}=8$. Vậy M = a + 2b - c = 0.