<s>[INST] Solution to a 4th order polynomial equation as Stillwell does (1989) I have a 4th order polynomial equation,
\begin{equation}
     x^4 + px^2 + qx + r = 0,
\end{equation}
which is of the same form as the beginning to a general solution for 4th order polynomials detailed by Stillwell's book, Mathematics and its history, First Edition (1989). Stillwell rewrites this as,
\begin{equation}
     \left(x^2 + p\right)^2 = px^2 - qx + p^2 - r
\end{equation}
and asserts that for any $y$,
\begin{equation}
     \left(x^2 + p + y\right)^2 = \left(p+2y\right)x^2 - qx + \left(p^2 - r + 2py + y^2\right).\tag{*}
\end{equation}
Then, he says that 

The quadratic $Ax^2 + Bx + C$ on the right-hand side will be a square
  if $B^2-4AC=0$, which is a cubic equation for $y$. We can therefore
  solve for $y$ and take the square root of both sides of the equation
  for $x$, which then becomes quadratic and hence also solvable. The final result is a formula for $x$ using just square and cube roots of rational functions of the coefficients.

I don't want to copy Stillwell's textbook verbatim, and I more-or-less understand his solution to the cubic equation, so I will just put the solution here, as it's not available online. Given a general cubic equation,
\begin{equation}
     x^3 + ax^2 + bx + c = 0,
\end{equation}
it holds true that a linear change of variable, $x = y - a/3$ gives
\begin{equation}
     y^3 = py + q,
\end{equation}
which is then solved by,
\begin{equation}
     y = \sqrt[3]{\frac{q}{2}+\sqrt{\left(\frac{q}{2}\right)^2-\left(\frac{p}{3}\right)^3}} + \sqrt[3]{\frac{q}{2}-\sqrt{\left(\frac{q}{2}\right)^2 - \left(\frac{p}{3}\right)^3}}.
\end{equation}
So, after doing all this, I still can't figure out how to recover a solution for $x$. However, I have a source (a part of a computer code) that claims the equation that must be solved is
\begin{equation}
     \left(x^2+p+y\right)^2 = ax^2+bx+c,
\end{equation}
which is of the same form as equation (*). My source does this by re-writing $ax^2+bx+c$ as a square, then "solving the equation" to obtain 2 results:
\begin{align}
     x^2 + \left(-\sqrt{a}\right)x + \left(-\frac{b}{2\sqrt{a}}+p+y\right) &= 0\\
     x^2 + \left(\sqrt{a}\right)x + \left(\frac{b}{2\sqrt{a}}+p+y\right) &= 0
\end{align}
If I can figure out the nitty-gritty details behind how these solutions came about, I would be set, but as it is I have spent several days grinding away at Stillwell's solutions without any progress. Please help!!
 [/INST] I'll try to add some details to what he says.
We can write
$$\left(x^2 + p + y\right)^2 = Ax^2+Bx +C\tag{*}$$
where $A=p+2y,B=-q,C=p^2-r+2py+y^2$.
The right-hand side will be a square $\left(\sqrt Ax+\frac{B}{2\sqrt A}\right)^2$ if $B^2-4AC=0$, i.e.
$$8y^3+20py^2+(16p^2-8r)y-q^2+4p^3-4pr=0$$
which is a cubic equation for $y$. 
We can therefore solve for $y$ and take the square root of both sides of the equation for $x$ 
$$\left(x^2 + p + y\right)^2 = \left(\sqrt Ax+\frac{B}{2\sqrt A}\right)^2$$
to have
$$x^2+p+y=\pm\left(\sqrt Ax+\frac{B}{2\sqrt A}\right)$$
which then becomes quadratic and hence also solvable.
</s>