The quartic equation: invariants and Euler’s solution revealed 1

RWD Nickalls 2

The Mathematical Gazette (2009); vol. 93 (March; No. 526), pp. 66–75 http://www.nickalls.org/dick/papers/maths/quartic2009.pdf

1 Introduction

The central role of the resolvent cubic in the solution of the quartic was first appreciated by Leonard Euler (1707–1783). Euler’s quartic solution first appeared as a brief section (§ 5) in a paper on roots of equations [1, 2], and was later expanded into a chapter entitled Of a new method of resolving equations of the fourth degree (§§ 773–783) in his Elements of algebra [3, 4]. Euler’s quartic solution was an important advance, in which he showed that each of the roots of√ a reduced√ quartic√ can be represented as the sum of three square roots, say ± 푟1 ± 푟2 ± 푟3, where the 푟푖 (푖 = 1, 2, 3) are the roots of a resolvent cubic. A quartic equation in 푥 is said to be reduced if the coefficient of 푥3 is zero. This can always be achieved by a simple change of variable. Motivated by the recent tercentenary of Euler’s birth, this article describes√ the geometric basis underlying both the 푟푖 and the sign of the product 푟1푟2푟3, these being two key aspects of Euler’s solution. Finally, we reveal the beautiful dynamic between Euler’s resolvent cubic and the quartic invariants 퐺, 퐻, 퐼, 퐽 [5, 6, 7], and propose a new class of algebraic object.

2 Geometric basis for the 푟푖

A significant property of the reduced quartic equation is that the four roots can be completely defined using only three parameters. For example, let 푧푗 (푗 = 1, 2, 3, 4) be the roots (see Figure 1) of a reduced quartic equation,

푍(푥) ≡ 푎푥4 + 푝푥2 + 푞푥 + 푟 = 0. (1)

As the sum of the roots is zero (the coefficient of the cubic term is zero), it follows that we can define the points midway between 푧1, 푧2 and 푧3, 푧4 as ±푔. Let 푧2 − 푧1 = 2훼 and 푧4 − 푧3 = 2훽. The four roots can then be expressed as follows: {︂ 푧1, 푧2 = −푔 ± 훼, 푧3, 푧4 = +푔 ± 훽. Since specifying one pair of quartic roots necessarily defines the remaining pair, there are just three different ways of allocating the pairs of roots, each associated with its own 푔, 훼, 훽, the inter-relationship between which lies at the heart of a remarkable symmetry which underpins the solution of the quartic. 1This minor revision of the original article corrects typographic errors and incorporates some explanatory footnotes. The original published version is available from the jstor archive at http://www.jstor.org/stable/40378672. 2Department of Anaesthesia, Nottingham University Hospitals, City Hospital Campus, Nottingham, UK. email: [email protected] RWD Nickalls Mathematical Gazette (2009); vol. 93, p. 66–75 2

gY g

...... α...... α β...... β ...... X z••••1 z2 z3 z4

Figure 1:

For example if, with no loss of generality, we let ⎧ ⎨ 푧3 + 푧4 = 2푔1, 푧3 + 푧1 = 2푔2, (2) ⎩ 푧3 + 푧2 = 2푔3, then 2(푔2 + 푔3) = 2푧3 + 푧1 + 푧2, = (푧1 + 푧2 + 푧3 + 푧4) + 푧3 − 푧4, = 푧3 − 푧4 = −2훽1, and similarly 2(푔2 − 푔3) = 푧1 − 푧2 = −2훼1,

and hence {︂ 훼1 = −(푔2 − 푔3), 훽1 = −(푔2 + 푔3).

Thus the 훼푘, 훽푘 (푘 = 1, 2, 3) are actually simple functions of the 푔푖 (푖 ≠ 푘) such that each of the four roots 푧푗 can be expressed as a function of the 푔푖 alone, as follows 3: ⎧ 푧1 = −푔1 − 훼1 = −푔1 + (푔2 − 푔3) = −푔1 + 푔2 − 푔3, ⎪ ⎨ 푧2 = −푔1 + 훼1 = −푔1 − (푔2 − 푔3) = −푔1 − 푔2 + 푔3, (3) 푧3 = +푔1 − 훽1 = +푔1 + (푔2 + 푔3) = +푔1 + 푔2 + 푔3, ⎪ ⎩ 푧4 = +푔1 + 훽1 = +푔1 − (푔2 + 푔3) = +푔1 − 푔2 − 푔3.

2 Thus Euler’s 푟푖 are the same as the 푔푖 .

3 Euler’s resolvent cubic

Using these observations we can reconstruct a given reduced quartic equation, say Equation 1, which then leads to a resolvent cubic and hence to the solution. Let the roots of 푍(푥) = 0 be −푔 ± 훼 and 푔 ± 훽 (Figure 1).

푍(푥) ≡ {푥 − (−푔 − 훼)}{푥 − (−푔 + 훼)}{푥 − (푔 − 훽)}{푥 − (푔 + 훽)} = 0.

Expanding and letting 퐴 = 푔2 − 훼2 and 퐵 = 푔2 − 훽2, gives

푥4 + (−4푔2 + 퐴 + 퐵)푥2 + (2푔)(퐵 − 퐴)푥 + 퐴퐵 = 0.

3 For a 3D version of Euler’s solution in which the ±푔푖 are associated with the mid-points of the six edges of a regular tetrahedron, see Fig 2 in Nickalls (2012), The quartic equation: alignment with an equivalent tetrahedron, Mathematical Gazette, 96, 49–55; http://www.nickalls.org/dick/papers/maths/tetrahedron2012.pdf. RWD Nickalls Mathematical Gazette (2009); vol. 93, p. 66–75 3

We can eliminate 훼, 훽 by first equating coefficients with the monic form of Equation 1 giving ⎧ 2 ⎨ 푝/푎 = −4푔 + 퐴 + 퐵, 푞/푎 = 2푔(퐵 − 퐴), ⎩ 푟/푎 = 퐴퐵, and then eliminating 퐴 and 퐵 (using the identity 4퐴퐵 = 2퐴 × 2퐵), which gen- erates a resolvent sextic in 푔, the roots of which are the six values ±푔1, ±푔2, ±푔3. The substitution 푔2 ↦→ 푥 then generates Euler’s original resolvent cubic [1, 2, 3, 4] 푝 (︂푝2 − 4푎푟 )︂ 푞2 푅(푥) ≡ 푥3 + 푥2 + 푥 − = 0, (4) 2푎 16푎2 64푎2 푟 푔2, 푔2, 푔2 whose roots 푖 are therefore 1 2 3. The four√ roots√ of the√ reduced quartic 푍(푥) = 0 are among the eight possible values of ± 푟1 ± 푟2 ± 푟3; but in order to determine which four they are we need a way of allocating the signs correctly. 푥4 − 푙푥2 − 푚푥 − 푛 Euler, using a monic quartic of the form√ = 0, says he resolved the sign problem by noting that 푟1푟2푟3 = 푚/8, as follows [3, § 773]: √ . . . But it is to be observed, that the product . . . 푟1푟2푟3, must be 푚/ 푚/ √equal√ to √8, and that if 8 be positive, the product of the terms 푟1, 푟2, 푟3, must likewise be positive; Unfortunately Euler did not elaborate further on this, but the key to under- standing the sign problem is not difficult to find, since from Equation 2we have

8푔1푔2푔3 = (푧3 + 푧4)(푧3 + 푧1)(푧3 + 푧2), 3 2 = 푧3 + 푧3 (푧1 + 푧2 + 푧4) + 푧3(푧2푧1 + 푧2푧4 + 푧1푧4) + 푧4푧1푧2.

Now 푧1 + 푧2 + 푧4 = −푧3 (since Σ푧푗 = 0), hence

8푔1푔2푔3 = 푧1푧2푧3 + 푧2푧3푧4 + 푧3푧4푧1 + 푧4푧1푧2, (5)

and so 8푔1푔2푔3 is actually one of the four elementary symmetric functions of the roots 푧푗. Its value is therefore equal to −1× the coefficient of the 푥-term of the monic form of the reduced quartic equation 푍(푥) = 0, and so we have √ 8 푟1푟2푟3 = 8푔1푔2푔3 = −푞/푎, (5푎) √ which is equivalent to Euler’s 푟1푟2푟3 = 푚/8. √ 4 Geometric basis for the sign of 푟1푟2푟3 A useful way of ‘seeing’ the quartic algebra at work is to express the coefficients

in terms of the key ‘visible’ parameters 휀, 푦푁푧, 푦푁푧′ shown in Figure 2, as follows: Let 퐹 (푋) be a quartic polynomial with real coefficients (푎 ≠ 0)

퐹 (푋) ≡ 푎푋4 + 푏푋3 + 푐푋2 + 푑푋 + 푒, (6) with invariants [6, p. 76] ⎧ 퐺 = 푏3 + 8푎2푑 − 4푎푏푐, ⎪ ⎨ 퐻 = 8푎푐 − 3푏2, (7) 퐼 = 12푎푒 − 3푏푑 + 푐2, ⎪ ⎩ 퐽 = 72푎푐푒 + 9푏푐푑 − 27푎푑2 − 27푒푏2 − 2푐3. RWD Nickalls Mathematical Gazette (2009); vol. 93, p. 66–75 4

. . . . Y . . . . . Z(x) ...... T2 ...... Nz ...... • •...... I ...... 1 ...... • ε ...... X ...... • N ...... T1 z′ ...... •...... ε .. I2 ...... Z′(x) . . •...... • T3

Figure 2: The reduced quartic 푍(푥), turning points (푇1, 푇2, 푇3), points of inflection ′ (퐼1, 퐼2), and first differential 푍 (푥). The 푥-coordinates of the points of

inflection are ±휀. The curves intersect the 푦-axis at points 푁푧 and 푁푧′ .

Let its reduced form 푍(푥) be generated by the translation 푋 ↦→ 푥 + 푋푁푓 , where 푋푁푓 = −푏/(4푎). Using Taylor’s theorem we have

퐹 ′′ 푋 4 ( 푁푓 ) 2 ′ 푍(푥) ≡ 퐹 (푥 + 푋푁 ) = 푎푥 + 푥 + 퐹 (푋푁 )푥 + 퐹 (푋푁 ). (8) 푓 2 푓 푓 ′ If 푍(푥) and 푍 (푥) intersect the 푦-axis in points 푁푧 and 푁푧′ respectively, then Equation 8 can be expressed as

4 2 2 푍(푥) ≡ 푎푥 − 6푎휀 푥 + 푦푁푧′ 푥 + 푦푁푧 (9) where (see Equation 4 and Figures 2, 3)

⎧ (3푏2 − 8푎푐) −퐻 −푝 휀2 ≡ ≡ , ⎪ = 2 2 ⎪ 48푎 48푎 6푎 ⎪ 2 ⎪ 퐼 3퐻 ⎨ 푦 = 퐹 (푋푁 ) ≡ − ≡ 푟, 푁푧 푓 12푎 482푎3 (10) ⎪ ′ 퐺 ⎪ 푦 ′ 퐹 푋 ≡ ≡ 푞, ⎪ 푁푧 = ( 푁푓 ) 2 ⎪ 8푎 ⎩⎪ 2 ′′ −12푎휀 = 퐹 (푋푁푓 ).

Expressing the reduced quartic 푍(푥) in this form (Equation 9) greatly facilitates visualisation, since we can now ‘see’ how the configuration of the curves 푍(푥) and 푍′(푥) is related to the coefficients. For example (assuming 푎 > 0), if the 푥2 term is positive then 휀 is complex (휀2 < 0), and so the quartic will have two complex points of inflection and hence only one real turning point (cf. [10]). RWD Nickalls Mathematical Gazette (2009); vol. 93, p. 66–75 5

If 푥푇푖 are the 푥-coordinates of the turning points of 푍(푥), then by differenti- ating Equation 9 we have (see Equations 5푎 and 10)

−푦푁푧′ √ 4푥푇 푥푇 푥푇 = = 8 푟1푟2푟3, (11) 1 2 3 푎 √ and hence the sign of 푟1푟2푟3 is the same as that of −푦 ′ /푎 and 푥푇 푥푇 푥푇 . It 푁푧 √ 1 2 3 follows, therefore, that we can actually ‘see’ the correct sign of 푟1푟2푟3 simply by observing the signs of the abscissae of the turning points of the reduced quartic, or by noting the location of 푁푧′ in relation to the abscissa. For example (assuming 푎 > 0), if the roots 푧푗 are such that the middle

turning point, 푇2, is to the left of the 푦-axis, then not only will 푦푁푧′ be negative 푥 (Figure 2) but just two of the three√ 푇푖 will be negative resulting in a positive product for 푥푇1푥푇2푥푇3, and hence 푟1푟2푟3 will also be positive (see Equation 11).

Conversely, if the middle turning point is to the right of the 푦-axis, then 푦푁푧′ will be positive, and only one of the 푥푇푖 will be negative making the product

푥푇1푥푇2푥푇3 negative.

5 Roots

푧 푍 푥 any As regards the roots 푗 of√ the reduced quartic ( ), we can initially choose√ sign combination for the 푟푖, and then evaluate the sign of the product 푟1푟2푟3.

If the sign of the product is the same as that of −푦푁푧′ /푎 (see Equation 11) then we have a valid combination of signs, and can proceed to determine the four 푧푗 one using Equation√ 3.√ Otherwise,√ it is only necessary to change the sign of any of the 푟푖 (say, 푟1 → − 푟1), and proceed as before using Equation 3. 푦 푥 When the reduced quartic is√ symmetric about the -axis one of the 푇푖 will be zero and hence the product 푟1푟2푟3 is zero. However, the solution in this

case is trivial since 푍(푥) is then an even function as 푦푁푧′ is also zero.

6 Application

Since all resolvent cubics of the quartic can be transformed to a standard form [9], typically expressed as [6, p. 77]

푇 (푥) ≡ 푥3 − 3퐼푥 + 퐽, (12) we can solve any quartic by solving instead a simple reduced form of the resolvent, say 푇 (푥) = 0, and then recover the roots of Euler’s resolvent using the transformation which carries the reduced form back to 푅(푥).

For example, the translation 푥 ↦→ 푥 + 푥푁푟 to reduce 푅(푥), for which 2 푥푁푟 = −푝/(6푎) ≡ 휀 , generates the reduced form 푆(푥), as follows: 퐼 퐽 푆(푥) ≡ 푅(푥 + 휀2) ≡ 푥3 − 푥 + . (13) 48푎2 1728푎3 The substitution 푥 ↦→ 푥/(12푎) then scales 1728푎3푆(푥) to 푇 (푥), and hence if the roots of 푆(푥) = 0 and 푇 (푥) = 0 are 푠푖 and 푡푖 respectively, then

2 푡푖 2 푟푖 = 푠푖 + 휀 = + 휀 . (14) 12푎 This convenient approach is illustrated in Example 1. RWD Nickalls Mathematical Gazette (2009); vol. 93, p. 66–75 6

.. .. Y ...... 2 H .. xN = ε = − . r .. ρ2 ...... R(x) ...... T ′...... T δr .. .. • ...... • ...... J .. X ...... P ... . yN = .. 2 ...... r ...... ρ3 .. G ..•...... y = − ...... P ...... ρ1 ...... •...... Nr ...... 3 ...... √ ...... 2 I ...... h = ...... r ...... ρ3 ...... • .. .. √I .. .. δr = .. ρ4 ..

Figure 3: Euler’s resolvent cubic 푅(푥) with three real roots (ℎ2 > 푦2 , i.e. 4퐼3 > 퐽 2) 푟 푁푟 which are all positive (휀2 > 0, 푥2 > 훿2). The conditions 휀2 > 0, 푥2 < 훿2 푁푟 푟 푁푟 푟 are associated with two negative roots (dashed curve). Note that 퐺2, 퐻, 퐼, 퐽 are constant multiples respectively of the resolvent’s geometric 2 2 6 2 3 parameters 푦푃 , 푥푁푟, 훿푟 , 푦푁푟 (휌1 = 64 푎 , 휌2 = 48푎 , 휌3 = 1728푎 , 휌4 = 12푎).

The invariants 퐼, 퐽 are readily visualised since any reduced cubic can be

expressed in terms of its geometric parameters 훿 and 푦푁 as in [8] 3 2 퐴푥 − 3퐴훿 푥 + 푦푁 = 0. (15) For example, equating coefficients between 푆(푥), 푇 (푥) and the monic form of Equation 15, and noting that ℎ2 = 4퐴2훿6 [8], shows that 퐼, 퐽 are simply constant 2 multiples of 훿 , 푦푁 as follows (Figure 3): ⎧ 퐴푟 = 퐴푠 = 퐴푡 = 1, ⎪ ⎪ 퐼 훿2 푎 2 훿2 푎 2 훿2, ⎪ = 푟 (12 ) = 푠 (12 ) = 푡 ⎨ 3 3 퐽 = 푦푁푟(12푎) = 푦푁푠(12푎) = 푦푁푡, (16) ⎪ ⎪ 3 (︂ )︂2 (︂ )︂2 (︂ )︂2 ⎪ 4퐼 ℎ푟 ℎ푠 ℎ푡 ⎩⎪ 2 = = = . 퐽 푦푁푟 푦푁푠 푦푁푡 Thus each of these invariants has a visible geometric interpretation in relation to Euler’s resolvent cubic, either as a position parameter with respect to the axes (퐺, 퐻, 퐽), or as a shape parameter (퐼). For example, we can now see that the condition 퐽 = 0 simply indicates that the 푁-point of the resolvent cubic lies on the 푥-axis and all that that implies (see Example 2). Similarly, the condition 퐼 = 0 indicates that the resolvent adopts the ‘cubic parabola’ form. Furthermore

푦푃 ≤ 0, which reveals how and why the resolvent cubic cannot have just a single negative root 4. The syzygy −27퐺2 = 퐻3 −48푎2퐼퐻 +64푎3퐽 [6, p. 76] is generated by substituting into 푆(푥) the coordinates of 푃 (퐻/(48푎2), −퐺2/(642푎6)). 4For real coefficients 퐺2 ≥ 0, and hence 푃 must always be on or below the 푥-axis. RWD Nickalls Mathematical Gazette (2009); vol. 93, p. 66–75 7

7 Euler’s cubic and the quartic root configurations

A very significant but seemingly overlooked aspect of Euler’s resolvent cubic is its beautiful and symmetric relationship with two important algebraic objects, namely the discriminant 4퐼3 − 퐽 2 and the seminvariant 퐻2 − 16푎2퐼, the signs of which distinguish between the various quartic root configurations [5, § 68; 6, p. 80; 7, p. 28]. Visualising the resolvent in relation to the invariants (Figure 3) reveals the mechanisms, as follows:

7.1 4퐼3 − 퐽 2 Since ℎ2 = 4퐴2훿6 [8], it follows from Equation 16 that −(4퐼3 − 퐽 2) = 푦2 − ℎ2. (17) 126푎6 푁푟 푟 퐼3 − 퐽 2 푦2 − ℎ2 Thus the quartic discriminant 4 is simply a constant multiple of 푁푟 푟, 푥 between 푦2 < ℎ2 on 푦2 ℎ2 the sign of which reflects whether the -axis lies ( 푁푟 푟), ( 푁푟 = 푟), outside 푦2 > ℎ2 or ( 푁푟 푟) the turning points of the resolvent cubic (Figure 3). 7.2 퐻2 − 16푎2퐼 The sign of this algebraic object distinguishes (when 휀2 > 0) between the then two possible quartic root configurations associated with the case 4퐼3 −퐽 2 > 0, namely (a) four real roots (퐻2 −16푎2퐼 > 0), and (b) four complex roots (퐻2 −16푎2퐼 < 0) [5, § 68]. Substituting for 퐻 (Equation 10) and 퐼 (Equation 16) gives 2 2 2 2 2 2 2 2 2 2 4 4 4 2 퐻 − 16푎 퐼 = (−48푎 휀 ) − 16푎 (12 푎 훿푟 ) = 3 4 푎 (휀 − 훿푟 ). 2 But 휀 = 푥푁푟 (Figure 3) and hence 퐻2 − 16푎2퐼 = 푥2 − 훿2. (18) 3244푎4 푁푟 푟 퐻2 − 푎2퐼 푥2 − 훿2 Thus 16 is just a constant multiple of 푁푟 푟 , the sign of which 휀2 > 푦 between 푥2 < 훿2 on 푥2 훿2 (when 0) reflects whether the -axis lies ( 푁푟 푟 ), ( 푁푟 = 푟 ), outside 푥2 > 훿2 or ( 푁푟 푟 ) the turning points of the resolvent cubic (cf. [6, p. 80, proposition 7]). For example (Figure 3), when a quartic with three real turning points (휀2 > 0) has four real roots (4퐼3 − 퐽 2 > 0) Euler’s cubic 푅(푥) has three positive real 푦 outside 푥2 > 훿2 roots—the -axis lies the two turning points—and so 푁푟 푟 and hence 퐻2 − 16푎2퐼 > 0. Conversely, when a quartic with three real turning points (휀2 > 0) has four complex roots (4퐼3 − 퐽 2 > 0), 푅(푥) then has exactly two negative real roots, 푇 ′ 푦 푥2 < 훿2 and so its turning point (Figure 3) lies to the left of the -axis ( 푁푟 푟 ), hence 퐻2 − 16푎2퐼 < 0.

7.3 A new class of object? Since 퐻2 − 16푎2퐼 functions with regard to the 푦-axis in exactly the same way that 4퐼3 − 퐽 2 functions with regard to the 푥-axis, I would like to suggest that this pair of algebraic objects should be regarded as forming a distinct class of object—thereby linking two previously independent algebraic quantities with a single unifying concept. RWD Nickalls Mathematical Gazette (2009); vol. 93, p. 66–75 8

8 Example 1

Solve 푓(푋) ≡ 푋4 − 11푋3 + 41푋2 − 61푋 + 30 = 0. ′ The key parameters are: 푎 = 1, 푋푁푓 = 11/4, 푌푁푓 ′ = 푓 (푋푁푓 ) = −15/8, 퐺 = −15, 퐼 = 28, 퐽 = −160, 휀2 = 35/48. Using say, 푇 (푥), we solve 5

푇 (푥) ≡ 푥3 − 84푥 − 160 = 0, √ the three 푡푖 being −8, −2, 10. The 푟푖 are therefore given by ⎧ √︂ √︂ √ 2 푡1 35 8 1 ⎪ 푟1 = 휀 + = − = , ⎪ 12푎 48 12 4 ⎪ ⎨⎪ √︂ √︂ √ 2 푡2 35 2 3 푟2 = 휀 + = − = , ⎪ 12푎 48 12 4 ⎪ ⎪ √︂ √︂ ⎪ √ 2 푡3 35 10 5 ⎩⎪ 푟3 = 휀 + = + = · 12푎 48 12 4

6 √ Since the sign of −푌푁푓 ′ /푎 is positive then the product of the 푟푖 must 푋 also be positive—which√ √ it√ is. Finally, adding 푁푓 recovers the quartic roots (푋푗 = 푋푁푓 ± 푟1 ± 푟2 ± 푟3) using (3) as follows:

⎧ 11 1 3 5 ⎪ 푋1 = − + − = 2, ⎪ 4 4 4 4 ⎪ ⎪ 11 1 3 5 ⎪ 푋2 = − − + = 3, ⎨ 4 4 4 4 11 1 3 5 ⎪ 푋3 = + + + = 5, ⎪ 4 4 4 4 ⎪ ⎪ 11 1 3 5 ⎩⎪ 푋4 = + − − = 1. 4 4 4 4

Even the solution of 푇 (푥) = 0 is greatly simplified since 훿, ℎ, 푦푁 are simple functions of 퐼 and 퐽 (see Equation 16). For example, 푇 (푥) = 0 has three real 2 2 3 roots in this case since (푦푁푡/ℎ푡) ≡ 퐽 /(4퐼 ) ≤ 1 [8].

9 Example 2

Explain the significance of 퐽 = 0, 퐼 > 0, for a quartic with four real roots. The condition 퐽 = 0 implies that Euler’s resolvent cubic has its 푁-point on the 푥-axis (Figure 3), and hence it has three roots in arithmetic progression. If also 퐼 > 0 (resolvent cubic has two real turning points), then the resolvent’s roots are distinct and (with the root at infinity) form a harmonic range. Since the roots of the parent quartic have the same cross-ratio they also form a harmonic range.

5 2 Note that we could instead solve 푆(푥) = 0, and then use 푟푖 = 휀 + 푠푖 (see Equation 14). 6Since 푌 ≡ 퐺/(8푎2) it is probably more convenient to use the sign of −퐺/푎 instead. 푁푓 ′ RWD Nickalls Mathematical Gazette (2009); vol. 93, p. 66–75 9

10 Acknowledgements

I would like to thank Professor JE Cremona (University of Warwick) for reading drafts of this paper and for his constructive comments. I also acknowledge the helpful suggestions of the anonymous reviewer.

11 References

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2 Bell J (2008). A conjecture on the forms of the roots of equations. arXiv:0806.1927v1 [math.HO]. http://arxiv.org/abs/0806.1927 [An English translation of Euler’s De formis radicum aequationum cujusque ordinis conjectatio (E30)]. 3 L. Euler. Vollst¨andigeAnleitung zur Algebra (Elements of algebra), 2 vols, Royal Academy of Sciences, St. Petersburg (1770). [Euler Archive, E387 (English): http://www.eulerarchive.org/docs/originals/E387e.P1S4.pdf] 4 C. R. Sangwin (Ed.). Euler’s Elements of Algebra. Tarquin Publications, St Albans, UK (2006). [English translation of Euler 1770 (E387)]

5 W. S. Burnside and A. W. Panton. The theory of equations: with an introduction to the theory of binary algebraic forms. (7th edn.) 2 vols; Longmans, Green and Co., London (1912). 6 J. E. Cremona. Reduction of binary cubic and quartic forms. J. Comput. Math., 2 (1999), pp. 62–92. http://www.lms.ac.uk/jcm/2/lms98007/ [The seminvariants 퐺 and 퐻2 −16푎2퐼 are denoted here by 푅 and 3푄 respectively] 7 P. J. Olver. Classical invariant theory. London Mathematical Society Student Texts No. 44. Cambridge University Press (1999). 8 R. W. D. Nickalls. A new approach to solving the cubic: Cardan’s solution revealed. Math. Gaz., 77 (Nov 1993) pp. 354–359. http://www.nickalls.org/dick/papers/maths/cubic1993.pdf. http://www.jstor.org/stable/3619777 9 R. S. Ball. Note on the algebraical solution of biquadratic equations. Quarterly Journal of Pure and Applied Mathematics, 7 (1866) pp. 6–9, 358– 369. http://www.nickalls.org/dick/papers/maths/ball1866quartic.pdf.

10 J. P. Dalton. On the graphical discrimination of the cubic and of the quartic. Math. Gaz., 17 (July 1933) pp. 189–196 http://www.jstor.org/stable/3607613

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