<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] 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:

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

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