Problem (Algebra):
Determine all values of $k$, with $k \neq 0$, for which the parabola

$$
y=k x^{2}+(5 k+3) x+(6 k+5)
$$

has its vertex on the $x$-axis.

Solution:
For the parabola to have its vertex on the $x$-axis, the equation

$$
y=k x^{2}+(5 k+3) x+(6 k+5)=0
$$

must have two equal real roots.

That is, its discriminant must equal 0 , and so

$$
\begin{aligned}
(5 k+3)^{2}-4 k(6 k+5) & =0 \\
25 k^{2}+30 k+9-24 k^{2}-20 k & =0 \\
k^{2}+10 k+9 & =0 \\
(k+1)(k+9) & =0
\end{aligned}
$$

Therefore, $k=-1$ or $k=-9$.

Final answer: $-1,-9$