Finding double root of $x^5-x+\alpha$ Given the polynomial
$$x^5-x+\alpha$$
Find a value of $\alpha>0$ for which the above polynomial has a double root.
Here's an animated plot of the roots as you change $\alpha$ from $0$ to $1$ I'm looking for $\alpha$ when the 2 points in the plot meet.

Also, this is not homework

As Daniel pointed out, since $p(x)$ will have a double root, $p'(x)$ must have the same root as well.
Also, by using Descartes rules of signs,
$$p(x) = x^5 -x +\alpha$$
$$p(-x) = -x^5 +x +\alpha$$
Therefore, p(x) has either 2 or 0 positive roots, 1 negative root, and either 2 or 4 complex root.
Since we are assumed there is a positive root, $p(x)$ will have 2 positive roots, 1 negative root, and 2 complex roots.
Taking the derivative,
$$p'(x) = 5x^4-1$$
Solving for x, we get that $$x =\left(\frac{1}{5}\right)^{\frac{1}{4}}$$
