Find the two values of $k$ for which $2x^3-9x^2+12x-k$ has a double real root. 
Find the two values of $k$ for which $2x^3-9x^2+12x-k$ has a double
  real root.

I've found one method which is to equate $$2x^3-9x^2+12x-k=2(x-r)^2(x-c)$$
Expanding and equating coefficients I get the system of equations :
\begin{array}
\space 2(c+2r) &=9 \\
2r(2c+r) &=12 \\
-2r^2c&=k \\
\end{array}
Solving this I've found the solutions $k=4,5$

However, I would like to know if there's some easier solution than the
  one I've already found, or  in general, if there's any other solution to the
  problem.

 A: Here is an easier way. If your polynomial $P_k(x) = 2x^3-9x^2+12x-k$ has a double real root $r$, then $P_k'(r) = 0$.
For such a root, you get $P_k'(r) = 6r^2-18r+12 = 0$, i.e. $r=1$ or $r=2$.
Then you solve $P_k(r)=0$ with respect to $k$. More precisely :


*

*If $r=1$, one needs $P_k(1)=5-k=0$ i.e. $k=5$.

*If $r=2$, one has $P_k(2)=4-k=0$ i.e. $k=4$.

A: You could find a relative maximum (or minimum) of f(x) = 2x3−9x2+12x, by considering, it´s derivative then choosing k accordingly. This way you don´need to find the roots explicitly.
A: Theoretical background.
In general if $\alpha$ is a multiple root of polynomial $f(x)$ then we can write: $$f(x)=(x-\alpha)^2g(x)$$
Taking the derivative on both sides we find:$$f'\left(x\right)=\left(x-\alpha\right)\left[2g\left(x\right)+\left(x-\alpha\right)g'\left(x\right)\right]$$
Evidently $f'(\alpha)=0$ so $\alpha$ appears to be a root of $f'\left(x\right)$ also.
A: From $2x^3-9x^2+12x-k=2(x-r)^2(x-c)$, using Vieta's formulas:
$$-\frac{-9}{2} = r + r + c \implies c = \frac{9}{2} - 2r$$
$$\frac{12}{2} = r \cdot r + r \cdot c + c \cdot r \implies6=r^2 + 2(9/2 - 2r)r$$
hence $6 = r^2 + 9r - 4r^2$ or $3(r^2 - 3r + 2) = 0 \implies r = 1, 2$. Using the remainder theorem, $2(1)^3 - 9(1)^2 + 12(1) - k = 0$ i.e $k =5$, or $2(2)^3 - 9(2)^2 + 12(2) - k = 0$ i.e $k =4$.
