Eight. For what value of $\mathrm{k}$, the roots of the equation $3\left(\mathrm{x}^{2}+3 \mathrm{x}+4\right)$ $=x(2-x)(2+k)$ are the sides of a right triangle

保持了源文本的换行和格式。
Eight, Solution: The original equation is rearranged as
$$
(\mathrm{k}+5) \mathrm{x}^{2}-(2 \mathrm{k}-5) \mathrm{x}+12=0 \text {. }
$$

Let the two roots of the equation be $x_{1}$ and $x_{2}$, then
$$
\begin{array}{l}
x_{1}+x_{2}=\frac{2 k-5}{k+5}, \\
x_{1} \cdot x_{2}=\frac{12}{k+5} .
\end{array}
$$

Since $x_{1}$ and $x_{2}$ are the cosines of the two acute angles of a right triangle, let $x_{1}=\cos \mathrm{A}, \mathrm{x}_{2}=\cos \mathrm{B}=\sin \mathrm{A}$, then
$$
x_{1}^{2}+x_{2}^{2}=\cos ^{2} A+\sin ^{2} A=1,
$$

which means $\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=1$;
$$
\begin{array}{l}
\therefore\left(\frac{2 k-5}{k+5}\right)^{2}-\frac{24}{k+5}=1, \\
(2 k-5)^{2}-24(k+5)=(k+5)^{2},
\end{array}
$$

Rearranging gives $\mathrm{k}^{2}-18 \mathrm{k}-40=0$,
$$
\therefore \mathrm{k}_{1}=-2, \mathrm{k}_{2}=20 \text {. }
$$
Since $\mathrm{x}_{1}$ and $\mathrm{x}_{2}$ are the cosines of two acute angles,
$$
\therefore \mathrm{x}_{1}>0, \quad \mathrm{x}_{2}>0 \text {. }
$$

But when $k=-2$, the original equation becomes $x^{2}+3 x+4=0$,
$$
\therefore x_{1}+x_{2}=-3 \text { and } x_{2}>0 \text {, which is a contradiction. }
$$

Therefore, $\mathrm{k}=-2$ does not meet the requirements and is discarded.
Finally, $\mathrm{k}=20$.