The Notebook Series The solution of cubic and quartic equations by R.S. Johnson Professor of Applied Mathematics School of Mathematics & Statistics University of Newcastle upon Tyne R.S.Johnson 006
CONTENTS List of equations Preface. 4 1. Introduction. 5. The solution of the cubic equation 7.1 Case D > 0 9. Case D = 0 10. Case D < 0 11.4 The trigonometric method 1 Exercises 1.. 1. The solution of the quartic equation. 14.1 Ferrari's method. 14. Euler's method.. 16 Exercises 18 Answers. 19 Index 0
List of Equations This is a list of the types of equation, and specific examples, whose solutions are discussed. z + az + bz+ c= 0 general solution. 7 y + βy+ γ = 0 general solution 7 x 6x 6= 0 9 4x 84x + 16 = 0.10 x 6x + 4= 0..1 y + βy+ γ = 0 trigonometric method..1 4 x + 4ax + 6bx + 4cx + d = 0 Ferrari s solution..14 4 x + 4x 1= 0..16 4 x + 4ax + 6bx + 4cx + d = 0 Euler s solution 16 4 z + 6αz + 4βz+ γ = 0 in Euler s solution 16
4 Preface This text is intended to provide an introduction to the methods for solving cubic and quartic equations. This topic is rarely covered in the degree programmes offered in the School of Mathematics & Statistics at Newcastle University, although the possibility of solving such equations is likely to be mentioned. The material has been written to provide a general introduction to the relevant ideas, and not as a text linked to a specific module which might need these results. Indeed, the intention is to present the material so that it can be used as an adjunct to a number of different modules or simply to help the reader gain a broader experience of mathematical ideas. The aim is to go beyond the methods and techniques that are presented in our modules, but standard ideas are discussed (and can be accessed through the comprehensive index). It is assumed that the reader has a basic knowledge of, and practical experience in, various aspects of elementary algebra, including some knowledge of complex numbers. This brief notebook does not attempt to include any applications of these equations; this is properly left to a specific module that might be offered in a conventional applied mathematics or engineering mathematics or physics programme. The approach adopted here is to present some general ideas, which might involve a notation, or a definition, or a theorem, or a classification, but most particularly methods of solution, explained with a number of carefully worked examples we present four. A small number of exercises, with answers, are also offered, although it must be emphasised that this notebook is not designed to be a comprehensive text in the conventional sense. Robin Johnson, September 006
5 1. Introduction The solution of algebraic equations, and by this we mean equations that involve x only in polynomial form, has held an exalted position in the history of mathematics with a pedigree that stretches back into the distant past. The simplest such equation is the linear equation ax + b = 0 where a ( 0 ) and b are given constants (in general complex-valued). Almost as simple, and certainly as familiar, is the quadratic equation ax + bx + c = 0 ( a 0). In this brief volume, we present the methods for solving cubic and quartic (sometimes called biquadratic) equations; however, we first should make a few general observations. The linear equation, with real coefficients, has a real solution; the quadratic equation, again with real coefficients, may have complex-valued solutions, as we know. Thus, in order to find all solutions of quadratic equations, we must admit complex numbers. The good news is that all equations, whether with real or complex coefficients, can be solved with no more than complex numbers: no additional new numbers need to be invented or defined. Furthermore, according to the fundamental theorem of algebra, all algebraic equations have at least one solution (usually called a root in this context). Let this root be z = z 0 we use z as a reminder that, in general, we expect complex roots then the equation f ()= z 0 can be written as f () z = ( z z0) g() z = 0. Then, for z z 0, the equation g()= z 0 also has at least one root, by the same theorem. This argument, continued until the remaining equation is linear, demonstrates that all polynomial equations, of degree n, have exactly n roots. (Note that some of these may be repeated roots.) Finally, the general solution (as a formula ) of the quadratic equation: F H 1 x = b± b 4ac a is very familiar; the aim of this Notebook is to develop corresponding results for cubic and quartic equations, but we can take this no further. A very deep and far-reaching result was proved, based on Galois theory, by Niels Abel in 184: it is impossible to find the roots of the general quintic equation (in the sense that no formula involving rational operations and root extractions exists). This problem of finding a solution I K,
had been attempted by many mathematicians over about 00 years, failure being put down to having not found the right trick ; Abel s proof explained the failure. Thus only equations of degrees 1,, and 4 can be solved in general; all the equations of higher degree, in general, cannot be solved. 6
7. The solution of the cubic equation We consider the general cubic equation z + az + bz+ c= 0, (1) which necessarily contains the cubic term, and for which we assume that c 0 (for otherwise the equation will have a root z = 0, leaving a quadratic equation to be solved for the other two roots). We describe the method that was published by Cardan in 1545, although there is some controversy over who should be regarded as the first to find this result. (It was probably first accomplished by Ferraro, who then passed the knowledge to Tartaglia, who then let Cardan into the secret, who in turn swore never to divulge the method!) The first stage is to set z = y+ α, to give y + ( a+ α) y + ( b+ aα + α ) y+ c+ bα + aα + α = 0, and then we choose α to remove the term in y i.e. α = a, so we obtain y + βy+ γ = 0, () where 1 β = b+ aα + α = b a, γ = c+ a 1 ab; the appearance of the 7 factor is simply a convenience. (Note that this transformation is unnecessary if a = 0.) The equation is immediately solvable (easily) if β = 0 (for then the solutions are the three cube roots of γ ); so hereafter, we assume that β 0. We also assume that γ 0, for otherwise the solution is again elementary: y = 0, y =± β. We comment, for future reference, that the case y = γ is conveniently expressed as Then the three solutions can be written as b g. y = 1 ( γ ) so that y = 1 1 γ 1 γ, ω γ, ω γ, where γ is any value of ( γ ) 1 (but it is simplest to choose the real value whenever γ is real, of course, which is the more common situation); here, 1, ω and ω are the three roots of unity i.e. iπ 1 ω = e = 1+ i i, ω d = e 4 i π = 1 d 1+ i i, with 1 + ω + ω = 0. () Note that the conjugate of ω, usually written ω, is ω, so that ω and ω form a complex pair. This special version of the cubic equation demonstrates that, for β = 0,
8 there is one real root, and a complex pair, when γ is real (although this same construction also holds when γ is complex-valued). We now turn to the problem of solving first we set y = u+ v to give y + βy+ γ = 0 ( β 0, γ 0 ); (4) ( u+ v) + β( u+ v) + γ = 0, but ( u+ v) = u + v + uv( u+ v) so y uvy = u + v. When we compare this with our original equation, (4), we see that we have a solution y u v = + if u + v = γ and uv = β. Thus we have u + v = γ and uv = β, which are the sum and product of two quantities, u and v, so these must be the roots of the quadratic equation w + γw β =0 i.e. w = 1 γ ± γ + 4β. RST UVW We now know both u and v ; let us select RST u 1 = γ + γ +4β UVW and then take the real root (in the case of γ and β real otherwise any root) for u as U, so The three roots for u are then RST UVW U = 1 γ + γ + 4β. (5) U, ωu, ω U = ( 1+ ω) U, (6)
9 1 d where ω = 1+ i i, as described earlier. Since uv = β, we have the corresponding roots for v: β U, β = ω β ω = + ω β β ( 1 ), = ω β U U U ω U U (7) (where we note that ω = 1 and 1+ ω + ω = 0). Hence we have the three roots of the cubic y + βy+ γ = 0, with y = u+ v from (6) and (7), written as U β, ωu + ( 1 + ω) β, ( 1+ ω) U ω β U U U. (8) However, although this is a complete description of the solution, it is only useful if D = γ + 4 β is real and non-negative (for then u is real for real γ ); further, the case D = 0 may lead, we might suppose, to some simplification. Thus we should consider separately the three cases D > 0, D = 0, D < 0, and hereafter, we will restrict the discussion to real γ and β..1 Case D > 0 With this choice, we have that RST U = 1 γ + γ + 4β UVW can be taken as a real number (positive or negative, but non-zero because β 0), so the other two roots are, necessarily, a complex pair: U β, β β + ω U + U U U which provides the complete solution. Example 1 F I HG K J F, ω + HG Find the roots of the cubic equation x 6x 6= 0. I K J β U U, U For this equation, we have β = and γ = 6, so γ + 4β = 6 = 4> 0; thus we take U = 4 and then β U = 4 =. The three roots are therefore
10 where ω = 1o 1+ i t.. Case D = 0 4 +, + 4 e jω, 4 4 e jω, We now have, from (5), U = γ, but this also gives v = γ because the roots of the quadratic for w are repeated. Thus uv = β becomes u = β > 0, where we note that 4β = γ < 0 so β < 0 ; hence we have u = v = ± β, and the sign to be selected must be that which is consistent with y + βy+ γ = 0 where y = u+ v = ± β. This clearly requires that u= v = (sgn γ) β. The three roots are therefore e j, u ω + ω = u u, u ω + ω = u e j, which shows perhaps not surprisingly that we have a repeated root of the cubic, and so all three roots are real. Indeed, the elementary structure in this case suggests that there may be a simpler way to solve this problem. Since the roots are u and u (repeated), we must have ( y+ u) ( y u) = y u y u y + βy+ γ ( = 0) so β = u and u = γ (both as we obtained above), with γ + 4β = 0. Thus with this latter condition applying, the roots are immediately selecting the real value of the cube root. γ and γ (repeated), Example Find the roots of the cubic equation 4x 84x + 16 = 0.
11 We have β = 4 49 and γ = 16 4, so γ b g b g i.e. D = 0. + 4β = 56 117649 + 4 64 117649 = 0 Thus we compute γ = 8 4 = 7, and this gives the roots 4 7, 7, 7.. Case D < 0 With D = γ + 4β < 0, the function U, RST U = 1 γ + γ + 4β, requires the cube root of a complex number. Now this can be found, of course, following conventional techniques, but it is not particularly straightforward. However, a rather surprising property of the solution exists even with D < 0 and we can take advantage of it. Write U = λ + i µ, then since u and v are the two roots RST w = 1 γ ± γ + 4β, we must have v = λ i µ since one must be the conjugate of the other. (Check: uv = λ + µ = 1 γ γ 4β 1 = β 4e j cubic are therefore λ + iµ + λ i µ, ωλ + µ + ω 1 b i g bλ iµ g, ωbλ + iµ g + ω bλ iµ g { }, which is correct.) The three roots of the b g b g i.e. λ, ωλ + i µ + ω λ iµ, ω λ + i µ + ω λ iµ or λ, ω λ + i µ + ω λ iµ UVW UVW b g b g b g b g, ω bλ + i µ g + ω bλ iµ g, so we have three real roots ( λ, λ µ, λ + µ ), and they are distinct. The Cardan approach necessarily requires us to deviate into the complex numbers in order to produce real roots! Because of this complication, this is often referred to as the irreducible case. There is, however, a much neater way of obtaining the roots. The cubic is y + βy+ γ = 0 with γ + 4β < 0,
1 so β < 0; write β = δ (δ > 0 ), and then set y = δ Y to give γ 4Y Y = δ δ where γ δ δ < 1, necessarily. Thus we may seek a solution Y = cosθ, for some θ, to produce 4cos θ cosθ = γ cosθ = δ δ, which leads directly to the three real roots for y. Example Find all the real roots of the cubic equation x 6x + 4= 0. In this case we have β = and γ = 4, so that γ x = cosθ, then we obtain 1 4cos θ cosθ = cosθ =, and so θ = π, 11π, 19 π,.... The three real roots are therefore 4 4 4 e j =, cos 11 cos 1 π 4 + 4β = 16 = 16< 0. Set e π 1 j, cos 19 1 e πj. This trigonometric approach that has been successful in the case D = γ + 4β < 0 can be applied more generally (although it cannot give all the solutions in a simple form)..4 The trigonometric method We are given y + βy+ γ = 0, and we introduce y = sgn( γ) β cosθ (where β is not necessarily negative; sgn denotes the signum function, giving the sign of γ ) to give e j i.e. cos β β 4cos θ cosθ = γ γ θ = ( β) β.
1 Of course, this recovers the result of. when β < 0 and γ ( β) β < 1. If γ + 4β = 0 (so β < 0 is still the case), then β = γ i.e. cosθ = 1. Thus θ = n π ( n = 01,, ) and so the three solutions are 4 sgn( γ) β, sgn( γ) which is the solution described in.. β, sgn( γ) β, This approach fails if D = γ + 4β > 0, for then we obtain cosθ > 1 which is impossible (for real θ ). However if we now introduce y = sgn( γ) β coshφ (for β < 0 ), then we obtain γ coshφ = ( β) β which leads directly to the real root that exists in this case. (For β > 0, we may use the corresponding transformation that replaces coshφ by sinhφ, but again we recover only the real solution directly.) Exercises 1 Find all the roots of these cubic equations: (a) x + 6x 0= 0; (b) x + 9x = 0; (c) x + 18x 6 = 0 ; (d) x x + = 0; (e) x 6x + = 0. **************** **********
14. The solution of the quartic equation We now consider the general quartic equation 4 x + 4ax + 6bx + 4cx + d = 0, (8) which is necessarily quartic by virtue of the term x 4. We insist that d 0 (for otherwise the equation reduces to a cubic and a fourth root x = 0 ); the inclusion of the factors 4 and 6 is merely an algebraic convenience. We shall present two versions of the method for solving this equation (although they eventually correspond very closely). The first is due to Ferrari (a student of Cardan) and the second was found by Euler. In each approach, it turns out that we must solve a suitable cubic equation (usually called the resolvent or reducing cubic of the quartic equation). That a cubic arises should be no surprise because, in the simple case where one root can be identified, the quartic will immediately factorise to produce cubic and linear factors..1 Ferrari s method The essential idea here is to introduce suitable terms (which cancel identically, of course) into the quartic expression 4 x + 4ax + 6bx + 4cx + d (9) so that this can be written as the difference of two squares. The equation will then take the form X Y = 0 and then X = ± Y are two quadratic equations each with two roots. In order to achieve this, we construct ex + ax+ b+ yj bλ x+ µ g e j b ge j b g e j = x + ax + b+ y x + ax + b+ y 4λ x + 4λµ x+ µ, (10) where y, λ and µ are yet to be chosen. Now we want (10) to be identical, for arbitrary x, to (9) (so that we are, equivalently, adding to (9) terms that cancel); this requires 4a + ( b+ y) 4λ = 6b; 4ab ( + y) 4λµ = 4c; ( b+ y) µ = d, and we note that the terms x 4 and x are automatically generated. These three equations yield λ = a + y b; λµ = ab ( + y) c; µ = ( b+ y) d which are consistent in the form bλµ g = λ µ if
15 ab ( + y) c = a + y b ( b+ y) d e j. (11) or 4y d + b 4ac y + bd + abc c a d b = 0 This is a cubic (already written in the standard form: no term in y ) which can be solved by the methods described in Chapter. Equation (11) is the resolvent (or reducing cubic) of the original quartic equation. (In passing, we comment that the constant term here can be expressed in a compact form: bd + abc c a d b = which demonstrates an underlying structure.) 1 a b a b c b c d, The quartic equation can now be written as e je j x + ax+ b+ y+ λ x+ µ x + ax+ b+ y λ x µ = 0 and so the problem is reduced to the elementary exercise of solving two quadratic equations: x + ( a+ λ ) x+ b+ µ + y = 0; x + ( a λ ) x+ b µ + y = 0. The solution of this pair generates the four solutions of the quartic, given suitable y, λ and µ. Any consistent choice is possible (and different choices will produce the same solutions of course but distributed differently through the two quadratic equations). To solve the quartic, therefore, the procedure is as follows: (a) select one solution of the cubic equation for y, from (11); (b) select a value of λ (or µ ) e.g. λ =+ a + y b ; ab ( + y) c (c) determine µ (or λ ) from λµ = ab ( + y) ce.g. µ = (and we a + y b must do this, rather than from µ = ( b+ y) d in order to ensure that we select the correct sign of µ ); (d) use these values of y, λ and µ (some of which may be complex) in the two quadratic equations and finally solve for x.
16 Example 4 Find all the roots of the quartic equation x 4 + 4x 1= 0. Here we have (see equation (8)): a = b= 0, c = 1, d = 1, so the resolvent is 4y + y 1= 0; this has a simple root y = 1 (we are lucky!). Thus we may use λ = 1 with solutions and µ = to produce the two quadratics x + x+ 1 = 0; x x+ 1+ = 0, 1 x = e ± 1 1 1 j, x = e ± + 1 i 1 j; these are the four roots of the quartic equation.. Euler s method Again, we start from equation (9): 4 x + 4ax + 6bx + 4cx + d = 0 with d 0 ; first we remove the x term by introducing z = x + a, to give 4 z + 6αz + 4βz+ γ = 0 (1) where 4. α = b a, β = c+ a ab, γ = d 4ac+ 6a b a The significant manoeuvre in this method is to seek a solution of equation (1) in the form z =± u ± v ± w, where all sign combinations are allowed (at this stage). This expression, when squared, yields d id i d id i d id i, z u v w= ± u ± v + ± u ± w + ± v ± w and when this is squared again:
17 z 4 ( u+ v+ w) z + ( u+ v+ w) { d id i d id i d id i} ( uv uw vw) d wid u v wi ( uv uw vw) d uid vid z. = 4 uv + uw + vw + u ± v ± w + v ± u ± w + w ± u ± v = 4 + + + 8 ± ± ± ± ± ± = 4 + + + 8 ± ± ± Thus we have generated a quartic with the z term absent: d id id i. 4 z ( u+ v+ w) z 8 ± u ± v ± w z+ ( u+ v+ w) 4( uv+ uw+ vw) = 0 Now this our equation (1) if d id id i α = ( u+ v+ w ); β = ± u ± v ± w and γ = ( u+ v+ w) 4( uv+ uw+ vw) = 9α 4 ( uv+ uw+ vw) ; i.e. ( u+ v+ w) = α ; uv + uw + vw = 1 4e 9α γj; uvw = 1 4 But these three terms are the coefficients (appropriately) of the cubic equation with roots u, v and w; this cubic is therefore e j. y y 1 y 1 + α + 9α γ β = 0 4 4 The standard procedure, as we know (Chap. ), is to remove the term in y by writing Y = y +α, which gives e j. 4Y α + γ Y+ αγ α β = 0 In terms of our original coefficients (see (9)), this becomes e j, 4Y d + b 4ac Y + bd + abc c a d b = 0 which is precisely the resolvent, equation (11), found in Ferrari s method. However, the construction of the solution follows a slightly different route in Euler s approach. From the three roots for Y, we obtain three values of y i.e. u, v and w; thus all the solutions are expressed as z =± u ± v ± w, given u, v and w. But β = d± uid± vid± wi, so we may write β.
18 β z =± u ± v d ± uid ± vi, and we need two solutions remember that Ferrari s method used only one of the resolvent. The four possible combinations of signs ( +, + ; +, ;, + ;, ) then generate the four solutions of the quartic equation. Because of the need to obtain two solutions of the reducing cubic (the resolvent) in Euler s method (one of which may be complex), Ferrari s approach is generally preferred: in this, we need find only a real root (and one always exists, if the coefficients of the quartic are real). Exercises Find all the roots of these quartic equations: 4 4 4 (a) x + x 5x + = 0; (b) x + x + x + 8x + = 0; (c) x + 5x + x + 8= 0; 4 4 (d) x 11x 6x + 10 = 0; (e) x + x + 5x + 5x + 1= 0. **************** **********
19 Answers Exercises 1 Throughout, we use ω = 1d 1+ i i (a cube root of unity). d i d i d i d i (a), 1 ω + 1+ ω, 1 ω + 1+ ω; (b) with u = 1+ 8, v = 8 1, then the three roots are u+ v, uω + vω, uω + vω ; (c) as in (b) with u = 18, v = 1 ; (d) approximately 1 769, 0 885± i 0 590 ; (e) approximately 60, 6, 0 40. Exercises 1 1 d i d i (a) 1±, 1± 5 d i; (b) 1 ± 5, 1± i 11 ; (c) 1 1± 15 1± 7 i, i ; (d) 1± 6, 1± ; (e) 1± i, 1d1± i 15i. **************** ********** d i d i 1
0 Index Abel 5 algebra fundamental thm 5 biquadratic equation 5 Cardan 7 case D=0 10 D>0 9 D<0 11 complex pair 7 cubic reducing 14, 15 cubic equation 7 D<0 11 D=0 10 D>0 9 degree higher than 4 6 equation biquadratic 5 cubic 7 linear 5 quadratic 5 quartic 14 quintic 5 Euler's method 16 Ferrari's method 14 Ferraro 7 fundamental theorem algebra 5 Galois 5 higher degree 6 linear equation 5 method Euler's 16 Ferrari's 14 trigonometric 1 quadratic equation 5 quartic equation 14 quintic equation 5 quintic equation Abel 5 reducing cubic 14, 15 resolvent 14,15,17 root 5 roots of unity 7 sgn 1 signum 1 Tartaglia 7 trigonometric method 1