已知 $4x^2 - \frac{x}{h} - 3 = 0$，其中 $h \ne 0$，证明不存在实数 $h$ 使得该方程具有相等的实根。 求正数常数 $k$ 的取值范围，使得直线 $y = kx$ 完全位于曲线 $y = -16x^2 + (2k - 1)x - 1$ 的上方。

1. Set up the discriminant condition for equal real roots.

For equal real roots, the discriminant $b^2 - 4ac = 0$.
Here, $a = 4$, $b = -\frac{1}{h}$, and $c = -3$.

2. Substitute the coefficients into the discriminant equation.

$\left(-\frac{1}{h}\right)^2 - 4(4)(-3) = 0$
$\frac{1}{h^2} + 48 = 0$

3. Show that the resulting equation has no real solutions for $h$.

$\frac{1}{h^2} = -48$
For any real value of $h \ne 0$, $h^2 > 0$, which means $\frac{1}{h^2} > 0$.
Since $\frac{1}{h^2}$ cannot be negative, there are no real values of $h$ that satisfy this equation. Thus, the equation cannot have equal real roots.

1. Set up the inequality for the line to lie completely above the curve

For the line $y = kx$ to lie completely above the curve $y = -16x^2 + (2k - 1)x - 1$, we must have:
$kx > -16x^2 + (2k - 1)x - 1$

2. Rearrange the inequality into standard quadratic form

$16x^2 + kx - (2k - 1)x + 1 > 0$
$16x^2 + (1 - k)x + 1 > 0$

3. Apply the discriminant condition for the quadratic to be always positive

For the quadratic expression to be always positive (since the coefficient of $x^2$ is $16 > 0$), the discriminant must be less than zero:
$b^2 - 4ac < 0$
$(1 - k)^2 - 4(16)(1) < 0$
$(1 - k)^2 - 64 < 0$

4. Solve the quadratic inequality for $k$ and apply the given condition

$k^2 - 2k + 1 - 64 < 0$
$k^2 - 2k - 63 < 0$
$(k - 9)(k + 7) < 0$
$-7 < k < 9$
Since $k$ is a positive constant ($k > 0$), the range of values is:
$0 < k < 9$