For what value of $k$ is one root of the equation 
For which value of $k$ is one root of the equation $x^2+3x-6=k(x-1)^2$ double the other?

My Attempt:
$$x^2+3x-6=k(x-1)^2$$
$$x^2+3x-6=k(x^2-2x+1)$$
$$x^2+3x-6=kx^2-2kx+k$$
$$(1-k)x^2+(3+2k)x-(6+k)=0$$

Now, $k\neq1$, $\Delta\geq0$ and since $x_1=2x_2$, we obtain:
$$3x_2=\frac{2k+3}{k-1}$$ and
$$2x_2^2=\frac{k+6}{k-1},$$
which gives
$$\frac{2(2k+3)^2}{9(k-1)^2}=\frac{k+6}{k-1}$$ or
$$k^2+21k-72=0,$$
which gives $k=3$ or $k=-24.$
Done!
