The latter may be solved as a cubic equation for~$k^2$. Any root $k^2 \neq 0$
gives a pair of quadratic factors of~$(21)$:
\[
\tag{23}
z^2 \pm 2kz + \tfrac{1}{2}q + 2k^2 \mp \frac{r}{4k}.
\]
The four roots of these two quadratic functions are the four roots of~$(21)$.
This method of Descartes (1596--1650) therefore succeeds unless every
root of~$(22)$ is zero, whence $q = s = r = 0$, so that $(12)$ is the trivial equation
$z^4 = 0$.

% [** F1: \small?]
For example, consider $z^4 - 3z^2 + 6z - 2 = 0$. Then $(22)$ becomes
\[
64k^6 - 3\cdot 32k^4 + 4\cdot 17k^2 - 36 = 0.
\]
The value $k^2 = 1$ gives the factors $z^2 + 2z - 1$, $z^2 - 2z + 2$. Equating these to zero, we
find the four roots $-1\pm \sqrt{2}$, $1\pm \sqrt{-1}$.

\paragraph{52. Symmetrical Form of Descartes' Solution.} To obtain this symmetrical
form, we use all three roots $k_1^2$, $k_2^2$, $k_3^2$ of~$(22)$.    Then
% It's fine to add spaces and newlines to a displayed equation to
% improve readability, but be sure never to leave a blank line within
% a displayed equation.
\[
k_1^2 + k_2^2 + k_3^2 = -\tfrac{1}{2}q,\qquad
k_1^2 k_2^2 k_3^2 = \frac{r^2}{64}.
\]
It is at our choice as to which square root of~$k_1^2$ is denoted by~$+k_1$ and
which by~$-k_1$, and likewise as to $\pm k_2$, $\pm k_3$. For our purposes any
choice of these signs is suitable provided the choice give
\[
\tag{24}
k_1 k_2 k_3 = -\frac{r}{8}.
\]

Let $k_1 \neq 0$. The quadratic function~(23) is zero for $k = k_1$ if
\[ % Alignment solely for readability.
(z \pm k_1)^2 = -\frac{q}{2} - k_1^2 \pm \frac{r}{4k_1}
              = k_2^2 + k_3^2 \mp \frac{8k_1k_2k_3}{4k_1}
              = (k_2 \mp k_3)^2.
\]
Hence the four roots of the quartic equation~$(21)$ are
\[ % Lining up parallel structures guards against typos.
\tag{25}
 k_1 + k_2 + k_3, \qquad
 k_1 - k_2 - k_3, \qquad
-k_1 + k_2 - k_3, \qquad
-k_1 - k_2 + k_3.
\]

% [** F1: Hard-coded formatting to match earlier formatters' work]
\begin{center}
\textbf{EXERCISES}
\end{center}

% [** F1: \small?]
\begin{itemize}
\item[1.]  Solve Exs.~4, 5 of §~48 by the method of Descartes.

\item[2.]  By writing $y_1$, $y_2$, $y_3$ for the roots $k_1^2$, $k_2^2$, $k_3^2$ of
\[
\tag{26}
64y^3 + 32qy^2 + 4(q^2-4s)y - r^2 = 0,
\]
show that the four roots of~$(21)$ are the values of
\[
\tag{27}
z = \sqrt{y_1} + \sqrt{y_2} + \sqrt{y_3} %[** F1: Missing period?]
\]
\end{itemize}
