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:
$$-\frac13\le k\le1$$

Suppose the equation has one root double the other. Then it can be written as
$$(x-a)(x-2a)=x^2-3ax+2a^2$$
Comparing coefficients between this expression and $x^2+(k+1)x+k^2$ we have
$$k+1=-3a\qquad2a^2=k^2$$
From the first equation we have $k=-3a-1$; substituting this into the second equation yields
$$2a^2=(-3a-1)^2=9a^2+6a+1$$
$$7a^2+6a+1=0$$
$$a=\frac{-6\pm\sqrt{6^2-4\cdot7\cdot1}}{2\cdot7}=\frac{-3\pm\sqrt2}7$$
From here we can recover the possible values of $k$:
$$k=-3\left(\frac{-3\pm\sqrt2}7\right)-1=\frac{2\pm3\sqrt2}7$$
