Let's determine the value of the parameter $k$ such that the roots of the following equation, when listed in ascending order, have the same difference between each pair of consecutive roots.

$$
4 x^{4}-(4 k+13) x^{2}+(9 k+9)=0
$$
I. solution. Our equation is quadratic in terms of $x^{2}$, and it can have at most two solutions for $x^{2}$. If these solutions are real, distinct, and positive, then we will get four roots for $x$, which are equal in absolute value in pairs. (1) can be rewritten as:

$$
x^{4}-\left(\frac{13}{4}+k\right) x^{2}+\frac{9}{4}(k+1)=x^{4}-\left(\frac{9}{4}+(k+1)\right) x^{2}+\frac{9}{4}(k+1)=0
$$

Noticing that the coefficient of $\left(-x^{2}\right)$ is exactly equal to the sum of the two factors of the constant term (and that the coefficient of $x^{4}$ is 1), it can be read off that the two solutions for $x^{2}$ are $9 / 4$ and $(k+1)$, so the equation is:

$$
4\left(x^{2}-\frac{9}{4}\right)\left\{x^{2}-(k+1)\right\}=0
$$

(This form can also be obtained naturally using the quadratic formula.) Thus, the roots of (1) are $\pm 3 / 2$ and $\pm \sqrt{k+1}$, provided that $k+1>0$ and $k+1 \neq 9 / 4$.

The roots can be listed in increasing order in two ways. If $k+1<9 / 4$, then $(-3 / 2)$ and $3 / 2$ are the 1st and 4th in the list:

$$
\begin{array}{lc}
x_{1}=-\sqrt{k+1}, & x_{2}=-\frac{3}{2}, \quad x_{3}=-x_{2}, \quad x_{4}=-x_{1} \\
x_{1}-3 x_{2}=0, & x_{1}=3 x_{2}=\frac{9}{2}, \quad k=\left(\frac{9}{2}\right)^{2}-1=\frac{77}{4}
\end{array}
$$

If, however, $k+1>9 / 4$, then $(-3 / 2)$ and $3 / 2$ are the 2nd and 3rd in the list:

$$
\begin{array}{lc}
x_{1}=-\sqrt{k+1}, & x_{2}=-\frac{3}{2}, \quad x_{3}=-x_{2}, \quad x_{4}=-x_{1} \\
x_{1}-3 x_{2}=0, & x_{1}=3 x_{2}=\frac{9}{2}, \quad k=\left(\frac{9}{2}\right)^{2}-1=\frac{77}{4}
\end{array}
$$

Accordingly, two values of $k$ satisfy: $-3 / 4$ and $77 / 4$.

II. solution. The squares of the roots of equation (1) are the roots of the

$$
4 y^{2}-(4 k+13) y+(9 k+9)=0
$$

quadratic equation, so we need to choose the value of the parameter $k$ such that (2) has two distinct positive roots. Let the smaller root of (2) be denoted by $a$, and the larger root by $b$. Then the roots of (1) in order of magnitude are

$$
x_{1}=-\sqrt{b}, \quad x_{2}=-\sqrt{a}, \quad x_{3}=-\sqrt{a}, \quad x_{4}=\sqrt{b}
$$

We need to choose the value of $k$ such that $a>0$ and

$$
x_{3}-x_{2}=x_{4}-x_{3}
$$

is satisfied (from which it already follows that $x_{2}-x_{1}=x_{3}-x_{2}$ is also satisfied), i.e., $3 \sqrt{a}=\sqrt{b}, b=9 a$ should hold. According to the relationship between the roots and coefficients, then for the coefficients of (2)

$$
\begin{gathered}
4 k+13=4(a+9 a) \\
9 k+9=4 \cdot 9 a^{2}
\end{gathered}
$$

is satisfied. According to the second, $k=4 a^{2}-1$, substituting this into the first, we get the quadratic equation for $a$:

$$
16 a^{2}+9=40 a
$$

whose roots are: $a_{1}=\frac{1}{4}, a_{2}=\frac{9}{4}$, so the sought values of $k$ are: $k_{1}=-\frac{3}{4}, k_{2}=\frac{77}{4}$. Since the values obtained for $a$ are positive, these values of $k$ satisfy the requirements of the problem.