Find the value of $k$ for which $kx^2 -5x-12 = 0$ has solutions $x=3$ and $ x = -\frac{4}{3}$.
So I can write two equations: $k(3)^2 - 5(3) - 12 = 0$ and $k(-\frac{4}{3})^2 - 5(-\frac{4}{3}) - 12 = 0$. Simplifying the first equation, I get $9k - 27 = 0$, so $k = 3$. Therefore, the value of $k$ for which $kx^2 -5x-12 = 0$ has solutions $x=3$ and $ x = -\frac{4}{3}$ is $k=3$. 