z4 + pz2 + qz + r = 0
for complex coefficients. The following solution method is due to
Euler [1729]. The basic idea is
to seek the roots as four plain linear combinations of three square roots.
Employing the ansatz
z = u + v + w yields
z2 = (u2 + v2 + w2) + 2(uv + uw + vw)
z4 = (u2 +
v2 + w2)2 +
4(u2 + v2 +
w2)(uv + uw + vw) +
4(u2v2 +
u2w2 +
v2w2) +
8uvw(u + v + w)
which after substitution into the original equation results in
(u2 + v2 +
w2)2 + p(u2 +
v2 + w2) +
(4(u2 + v2 + w2) +
2p)(uv + uw + vw) +
4(u2v2 +
u2w2 +
v2w2) +
(8uvw + q)(u + v + w) +
r = 0
By first requiring
(1) 4(u2 + v2 + w2) + 2p = 0
(3) 8uvw + q = 0
and subsequently exploiting (1), the remaining equation becomes
(2) 0 = (u2 + v2 +
w2)2 + p(u2 +
v2 + w2) +
4(u2v2 +
u2w2 +
v2w2) + r
=
(−p/2)2 + p(−p/2) +
4(u2v2 +
u2w2 +
v2w2) + r
=
4(u2v2 +
u2w2 +
v2w2) −
p2/4 + r
Introducing t1 = u2, t2 = v2, t3 = w2, equations (1–3) are reformulated as
(1.2)
t1 + t2 + t3 =
−p/2
(2.2)
t1t2 +
t1t3 +
t2t3 =
p2/16 − r/4
(3.2)
t1t2t3 =
q2/64
According to the Viète formulas, t1, t2, t3 are the roots to the cubic resolvent
(4) t3 + (p/2)t2 + (p2/16 − r/4)t − q2/64 = 0
Thus, the following possibilities arise
u = ∓√t1,
v = ∓√t2,
w = ∓√t3 from which eight linear
combinations for the ansatz (u + v + w)
could be formed. Regardless of the choices of sign,
the product uvw becomes either
√t1√t2√t3 or
-√t1√t2√t3.
Consequently, it is convenient to choose
u = √t1, v = √t2, w = √t3
and use (3) as the discriminator between case 1
(8uvw + q = 0) or case 2
(-8uvw + q = 0).
For case 1, the appropriate linear combinations are
u+v+w, u-v-w,
-u+v-w, -u-v+w
(each product of the three terms is uvw). For case 2,
the appropriate linear combinations are
-u-v-w, -u+v+w,
u-v+w, u+v-w
(each product of the three terms is -uvw).
For the case q ≠ 0, the cubic resolvent (4) is depressed by the change of variable t = s − p/6 into
s3 + 3bs − 2d = 0
where
b = −r/12 −
p2/144
d = q2/128 −
pr/48 + p3/1728
the roots of which yield
u = √(s1 − p/6), v = √(s2 − p/6), w = √(s3 − p/6)
The roots to the original quartic equation become
if |8uvw + q| < |8uvw − q|
z1 = u +
v + w
z2 = u −
v − w
z3 = −u +
v − w
z4 = −u −
v + w
else
z1 = −u −
v − w
z2 = −u +
v + w
z3 = u −
v + w
z4 = u +
v − w
For the biquadratic case q = 0, the cubic resolvent (4) is factorized into
(t2 + (p/2)t + p2/16 − r/4)t = 0
the roots of which are
t1 = −p/4 +
√(r/4)
t2 = −p/4 −
√(r/4)
t3 = 0
yielding w = 0 and
u = √(−p + 2√r)/2, v = √(−p − 2√r)/2
The roots to the original quartic equation simply become
z1 = u + v
z2 = u − v
z3 = −u + v
z4 = −u − v