I have a depressed quartic polynomial with three free parameters in the real numbers:
$x^{4}+qx^{2}+rx+s$
Furthermore, the discriminant is constrained to be zero and there are four real roots, exactly two of which are equal. I am only interested in the double root.
Wikipedia has the general solution. However, is there a simpler formula for this special case?
On
When a polynomial $f(x)$ has a double root $\alpha$, the factor $x-\alpha$ appears in both $f(x)$ and in its derivative $f'(x)$. And conversely, every common factor of $f$ and $f'$ is a double (or higher-order) repeated root. Thus, using the Euclidean algorithm for polynomials, applied to $f$ and $f'$, will tell all repeated roots and their multiplicities (the multiplicities are correct for any field not of characteristic $p>0$... which, if it doesn't mean anything to you, is almost surely irrelevant to your issue here).
On
Here's a quick way to get another formula (of sorts) for the double root.
A double root of a polynomial $P(x)$ is also a root of the derivative, $P'(x)$. So if $x$ is the double root of a depressed quadratic $P(x)=x^4+qx^2+rx+s$, then
$$\begin{align} x^4+qx^2+rx+s&=0\\ 4x^3+2qx+r&=0 \end{align}$$
Multiplying the first equation by $4$ and subtracting the second multiplied by $x$ leaves
$$2qx^2+3rx+4s=0$$
so $x$ is one of the two roots of this quadratic, i.e.,
$$x={-3r+\sqrt{9r^2-32qs}\over4q}\quad\text{or}\quad x={-3r-\sqrt{9r^2-32qs}\over4q}$$
What's left unclear, to me at least, is how to tell (in advance, just from the coefficients $q$, $r$, and $s$) which expression to use for a given depressed polynomial. To give an example, suppose $P(x)=(x-1)^2(x^2+2x-1)=x^4-4x^2+4x-1$. Then
$$x={-12\pm\sqrt{144-128}\over-16}={-12\pm4\over-16}$$
and the $-4$ gives the right answer. On the other hand, if $P(x)=(x+1)^2(x^2-2x-1)=x^4-4x^2-4x-1$, then
$$x={12\pm\sqrt{144-128}\over-16}={12\pm4\over-16}$$
and the $+4$ gives the right answer.
Yes, there is. You know that your roots (including multiplicities) can be expressed as $\{a, a, -a+δ, -a-δ\}$ (because the sum is zero after the normalization). Expand
$$(x-a)^2(x+a-δ)(x+a+δ) = x^4 - (2a^2 + δ^2)x^2 + 2aδ^2 x + a^4-δ^2a^2$$
and compare with your form. Next, note that the equations
$$\begin{aligned} - (2a^2 + δ^2) &= q \\ a^4-δ^2a^2 &= s \end{aligned}$$
can be both expressed using
$$A = a^2, D = δ^2$$
and they only lead to a quadratic equation for $A$. It's easy from there. You'll get $a^2$ but the sign of $a$ is correlated with the sign of the linear term (which otherwise only contains factors of a 2 and a square).
Generally:
$$\bbox[7px,border:2px solid]{a = (\mathop{\rm sgn} r)\cdot \sqrt{\frac{-q \pm \sqrt{q^2 + 12s}}6}}\ .$$
The sign inside the square root needs to be chosen so that $2aD = -2a(q+2a^2) = r$. For $s > 0$ only + guarantees a real $a$ but if $s < 0$ then both signs lead to a real solution and one of them corresponds to a quartic with a different $r$.
Example:
$$\begin{aligned} p(x) &= x^4 - 19x^2 + 6x + 72 \\ ⇒ 2A + D &= 19 \\ A^2 - AD &= 72 \\ ⇒ A^2 - A(19-2A) = 3A^2 - 19A &= 72 \\ ⇒ A &= \frac{19 \pm \sqrt{1225}}{6} = \frac{19 \pm 35}6, \quad D = 19-2A \\ A ≥ 0 ⇒ A &= 9, \quad D = 1 \\ ⇒ a &= 3, d = 1 ⇒ p(x) = (x-3)^2 (x+2) (x+4) \end{aligned}$$