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

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

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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!

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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 $x_2 = 2x_1$.

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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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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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