The latter may be solved as a cubic equation for k^2. Any root k^2 $$ 0
gives a pair of quadratic factors of (21):

(23)                      z^2 $$ 2kz + 1/2q + 2k^2 $$ 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.

For example, consider z^4-3z^2+6z-2 = 0.    Then (22) becomes

64k^6- 3·32k^4 + 4·17k^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$$ $$2, 1$$ $$-1.

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

k_1^2+k_2^2+k_3^2 = -1/2q,         k_1^2k_2^2k_3^2 = 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 $$k_2, $$k_3. For our purposes any
choice of these signs is suitable provided the choice give

(24)                    k_1k_2k_3 = -r/8.

Let k_1 $$ 0.    The quadratic function (23) is zero for k = k_1 if

(z$$k_1)^2 = -q/2 - k_1^2 $$ r/4k_1 = k_2^2 + k_3^2 $$ 8k_1k_2k_3/4k_1 = (k_2 $$ k_3)^2.

Hence the four roots of the quartic equation (21) are

(25)   k_1+k_2+k_3,      k_1-k_2-k_3,      -k_1+k_2-k_3,      -k_1-k_2+k_3.

EXERCISES

1.  Solve Exs. 4, 5 of § 48 by the method of Descartes.

2.  By writing y_1, y_2, y_3 for the roots k_1^2, k_2^2, k_3^2 of

(26)                   64y^3+32qy^2+4(q^2-4s)y-r^2 = 0,

show that the four roots of (21) are the values of

(27)                   z = $$y_1 + $$y_2 + $$y_3 
