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:
$$x^2+3x-6=k(x-1)^2$$
$$x^2+3x-6=k(x^2-2x+1)$$
$$x^2+3x-6=kx^2-2kx+k$$
$$(1-k)x^2+(3+2k)x-(6+k)=0$$
 A: You're almost there.
Just consider the two distinct cases where $k$ equals $1$ or not. If $k=1$, your polynomial is of degree one and only has one root, so $1$ is not a solution.
If $k \not=1$, use the classical https://en.wikipedia.org/wiki/Quadratic_formula to find the roots $x_1,x_2$ (for some values of $k$, they may be complex) and then solve for $x_1 = 2x_2$ or $x_2 = 2x_1$.
A: 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!
A: Hint: Let $a$ and $2a$ be the two roots.
Then
$$
-\dfrac{3+2k}{1-k}=3a,
\qquad
-\dfrac{6+k}{1-k}=2a^2
$$
A: If $ax^2+bx+c$ has two roots, then the sum of the roots is equal to $-b/a$ and the product of the roots is equal to $c/a$. In your case, if one root is $p$, then the other one would be $2p$; applying the formulas to:

$$(1-k)x^2+(3+2k)x-(6+k)=0$$

$$\frac{3+2k}{k-1}=3p \quad\mbox{ and }\quad \frac{6+k}{k-1}=2p^2$$
