Finding double root of $x^5-x+\alpha$

Question

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

Answer 1

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

$$\alpha=\frac{4}{5\sqrt[4]{5}}$$

Answer 2

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 $$x =\left(\frac{1}{5}\right)^{\frac{1}{4}}$$

Answer 3

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 $\alpha^4={256\over 3125}$.

Answer 4

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 $\alpha$

$$3125\alpha^4-256=0$$ $$3125\alpha^4=256$$ $$\alpha^4=\frac{256}{3125}$$ $$\alpha=\frac{4}{\sqrt[4]{3125}}$$ $$\alpha=\frac{4}{5\sqrt[4]{5}}$$