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
 A: A multiple root is a common root of both $f(x)$ and $f^{\prime}(x)$.
$$f^{\prime}(x)=5x^4-1$$
and if you are interested only in real roots (i get this impression that you are)
then $$x=\frac{1}{\sqrt[4]{5}}$$
and further we must have 
$$f\left(\frac{1}{\sqrt[4]{5}}\right)=0$$  this gives
$$\alpha=\frac{4}{5\sqrt[4]{5}}$$
A: If you write a polynomial $p(x)=(x-r_1)^{n_1}\ldots (x-r_k)^{n_k}$ where we assume each $r_i\ne r_j$ when $i\ne j$ and $n_i\ge 1$ for $1\le i\le k$.
Then if we take derivatives, the product rule gives us
$$p'(x)=\sum_{i=1}^k n_i(x-r_i)^{n_i-1}\prod_{i\ne j}(x-r_j)^{n_j}$$
If $n_m=1$ for some $1\le m\le k$, then we get that $(x-r_m)$ divides each term except one, hence by the rules of divisibility, it cannot divide the sum, hence the derivative does not share this particular root. On the other hand, if $n_m>1$, then $(x-r_m)$ divides every term in that sum, hence the two share that root.
So the common roots of $p(x)$ and $p'(x)$ are exactly the multiple roots of $p$.
As Peter notes, for this case this implies that
$$(x^5-x+\alpha, 5x^4-1)=(5x^4-1, 4x-5\alpha)=(4x-5\alpha, {3125\over 256}\alpha^4-1)$$
so $\alpha^4={256\over 3125}$.
A: Somebody posted a hint answer which made solving it easy, but they quickly deleted it. Here's a solution using that answer.
When the discriminant of a polynomial is zero, atleast two roots coincide.
The discriminant of $x^5-x+\alpha$ is $3125\alpha^4-256$ solving for $\alpha$
$$3125\alpha^4-256=0$$
$$3125\alpha^4=256$$
$$\alpha^4=\frac{256}{3125}$$
$$\alpha=\frac{4}{\sqrt[4]{3125}}$$
$$\alpha=\frac{4}{5\sqrt[4]{5}}$$
A: 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 $$x =\left(\frac{1}{5}\right)^{\frac{1}{4}}$$
