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:
$$-\frac13\le k\le1$$
 A: 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 $k=\frac{3}{7}\sqrt{2}\pm\frac{2}{7}$.
A: 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 $k$:
$$k=-3\left(\frac{-3\pm\sqrt2}7\right)-1=\frac{2\pm3\sqrt2}7$$
A: Let $x_1$ and $x_2$ be the two roots and $x_1 = 2x_2$. Then we have by Vieta's formulae: $-(k+1_ = x_1 + x_2 = 3x_2 \implies x_2 = -\frac{1+k}{3}$. On the other hand: $k^2 = x_1x_2 = 2x_2^2 \implies x_2 = \pm \frac{k}{\sqrt{2}}$
Now equate the two equation and you will be able to solve for $k$ easily..
A: Let the given polynomial be, $f(x)$ and the roots be, $\alpha, 2\alpha.$ 
Then, $f(x)= x^2-(\alpha + 2\alpha)+\alpha \cdot 2\alpha =x^2-3\alpha+2\alpha ^2$ (How?)
$\therefore k^2=2\alpha ^2 $
$\implies k=\pm \sqrt{2}\alpha ...(1)$
Again, $k+1=-3\alpha...(2)$
Dividing (2) by (1), we get
$$\frac{k+1}{k}=\frac{-3}{\pm\sqrt{2}}$$
The solution of the equation is left to you. I hope it is easy to continue from here.
