Step 1: Use the equal-roots property instead of the discriminant.
When a quadratic $ax^2+bx+c=0$ has real and equal roots, both roots are equal to $-\frac{b}{2a}$, and their product equals $\frac{c}{a}$. Here $a=k$, $b=-6$, $c=-4$.
Step 2: Write the repeated root and the product condition.
The repeated root is $\alpha = -\frac{-6}{2k} = \frac{3}{k}$. Since both roots equal $\alpha$, their product is $\alpha^2$, which must equal $\frac{c}{a} = -\frac{4}{k}$.
Step 3: Solve for k.
\[ \left(\frac{3}{k}\right)^2 = -\frac{4}{k} \] \[ \frac{9}{k^2} = -\frac{4}{k} \] Multiplying both sides by $k^2$ (since $k \neq 0$): \[ 9 = -4k \] \[ k = -\frac{9}{4} \]
\[ \boxed{k = -\frac{9}{4}} \]