<s>[INST] 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:
$$x^2+3x-6=k(x-1)^2$$
$$x^2+3x-6=k(x^2-2x+1)$$
$$x^2+3x-6=kx^2-2kx+k$$
$$(1-k)x^2+(3+2k)x-(6+k)=0$$
 [/INST] Hint: Let $a$ and $2a$ be the two roots.
Then
$$
-\dfrac{3+2k}{1-k}=3a,
\qquad
-\dfrac{6+k}{1-k}=2a^2
$$
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