<s>[INST] Real values of $k$ for which $x^2+(k+1)x+k^2$ has one root double the other 
For what real values of $k$ does $x^2+(k+1)x+k^2$ have one root double the other?

For a start, I found the range of $k$ which endows this equation with real roots:
$$-\frac13\le k\le1$$
 [/INST] Let $a$ and $b$ be the roots. Since 
$$
(x-a)(x-b)=x^2+(k+1)x+k^2\\
x^2-(a+b)x+ab=x^2+(k+1)x+k^2
$$
it follows (by comparing corresponding coefficients) that 
$$
\begin{cases}
a+b=-(k+1) & (1)\\
ab=k^2 & (2)
\end{cases}
$$
If $a=2b$, then
$$
\begin{cases}
3b=-(k+1) & (3)\\
2b^2=k^2 & (4)
\end{cases}
$$
Flugging (3) inside (4) gives 
$$
\sqrt{2}\left|-\frac{k+1}{3}\right|=|k|
$$
so $k=\frac{3}{7}\sqrt{2}\pm\frac{2}{7}$.
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