6 . For the real-coefficient quadratic equation $2 \mathrm{kx}^{2}+(8 \mathrm{k}+1) \mathrm{x}+8 \mathrm{k}=0$ to have two distinct real roots, the range of $\mathrm{k}$ is:
(A) $\mathrm{k}-\frac{1}{16}$ s
(C) $k \geqslant-\frac{1}{16}$,
(D) None of the above results are correct.
Answer: (D).
Solution 1: The condition for the equation to have two unequal real roots is $\Delta>0 \quad$ and $k \neq 0$.
That is, $\left\{\begin{array}{l}(8 k+1)^{2}-4 \cdot 2 k \cdot 8 k>0, \\ k \neq 0 .\end{array}\right.$
Solving this, we get $k>-\frac{1}{16}$ and $k \neq 0$.
Therefore, options A, B, and C are incorrect.
Solution 2: $\because$ The equation is a quadratic equation, $\quad \therefore \mathrm{k} \neq 0$.
Options (B) and (C) clearly do not satisfy the condition $k \neq 0$. We only need to consider (A) and (D).
Let's assume $\mathrm{k}=-\frac{1}{8}\left(-\frac{1}{8}<-\frac{1}{16}\right)$, and substitute it into the original equation.

We get $-\frac{1}{4} x^{2}-1=0$
This equation has no real solutions. Therefore, (A) is incorrect.
Since the question states that only one answer is correct, the correct answer is (D).