5.2.1 * Given the real-coefficient equation
$$
x^{3}+2(k-1) x^{2}+9 x+5(k-1)=0
$$

has a complex root with a modulus of $\sqrt{5}$. Find the value of $k$, and solve this equation.
By the principle of conjugate pairs, we know that (1) has one real root and two complex roots with a modulus of $\sqrt{5}$, which are conjugates of each other. Let these three roots be $a+b \mathrm{i}, a-b \mathrm{i}, c(a, b, c \in \mathbf{R})$, then
$$
a^{2}+b^{2}=5
$$

By Vieta's formulas, we have $\left\{\begin{array}{l}(a+b \mathrm{i})+(a-b \mathrm{i})+c=-2(k-1), \\ (a+b \mathrm{i})(a-b \mathrm{i})+c(a+b \mathrm{i})+c(a-b \mathrm{i})=9, \\ (a+b \mathrm{i})(a-b \mathrm{i}) c=-5(k-1) .\end{array}\right.$
Combining (2), we get $\left\{\begin{array}{l}2 a+c=-2(k-1), \\ a c=2, \\ c=-k+1 .\end{array}\right.$
From (3) and (5), we know $c=1-k, a=\frac{1}{2}(1-k)$, and substituting these into (2), we get $k=-1$ or 3. Solving equation (1) further, we find: when $k=-1$, the solutions of (1) are $1+2 \mathrm{i}, 1-2 \mathrm{i}, 2$; when $k=3$, the solutions of (1) are $-2,-1+2 \mathrm{i},-1-2 \mathrm{i}$.