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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 2006

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CONTENTS

List of equations ……… 3

Preface …………. ……… 4

1. Introduction ………. 5

2. The solution of the cubic equation ……… 7

2.1 Case D > 0 ……… 9

2.2 Case D = 0 ………10

2.3 Case D < 0 ………11

2.4 The trigonometric method ……… 12

Exercises 1 ……….. 13

3. The solution of the quartic equation ………. 14

3.1 Ferrari's method ………. 14

3.2 Euler's method ……….. 16

Exercises 2 ……… 18

Answers ………. 19

Index ……… 20

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List of Equations

This is a list of the types of equation, and specific examples, whose solutions are discussed.

z3+az2+bz c+ =0 – general solution………. 7

y3+3βy+ =γ 0 – general solution………7

x36x− =6 0 ………9

343x384x+16 0= ……….10

x36x+ =4 0 ………..12

y3+3βy+ =γ 0 – trigonometric method………..12

x4+4ax3+6bx2 +4cx d+ =0 – Ferrari’s solution………..14

x4+4x− =1 0 ………..16

x4+4ax3+6bx2 +4cx d+ =0 – Euler’s solution………16

z4+6αz2+4β γz+ =0 – in Euler’s solution………16

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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 2006

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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

ax2+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 z g z0) ( )=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:

x= 21a

FH

− ±b b24ac

IK

,

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 1824: 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 –

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had been attempted by many mathematicians over about 200 years, failure being put down to having not found the right ‘trick’; Abel’s proof explained the failure. Thus only equations of degrees 1, 2, 3 and 4 can be solved in general; all the equations of higher degree, in general, cannot be solved.

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2. The solution of the cubic equation

We consider the general cubic equation

z3+az2+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

y3+(a+3α)y2+ +(b 2aα+3α2)y c b+ + α+aα2+α3 =0, and then we choose α to remove the term in y2 i.e. α = − a 3 , so we obtain

y3+3βy+ =γ 0, (2) where 3β = +b 2aα+3α2 = −b 13a2, γ = +c 272 a313ab; the appearance of the factor ‘3’ 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= ± −3β . We comment, for future reference, that the case y3= −γ is conveniently expressed as

y3 = × −1 ( γ) so that y=11 3

b g

γ 1 3. Then the three solutions can be written as

γ

3 , ω 3 γ , ω2 3− , γ

where 3 γ is any value of (γ)1 3 (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

ω2 are the three roots of unity i.e.

ω =e2iπ 3 =21

d

− +1 i 3

i

, ω2 =e4iπ 3= −12

d

1+i 3

i

, with 1+ +ω ω2 =0. (3) Note that the conjugate of ω, usually written ω , is ω2, so that ω and ω2 form a complex pair. This special version of the cubic equation demonstrates that, for β= 0,

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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

y3+3βy+ =γ 0 (β ≠ 0, γ ≠ 0 ); (4) first we set y u v= + to give

(u v+ )3+3β(u v+ + =) γ 0, but (u v+ )3 =u3+v3+3uv u v( + ) so y33uvy u= 3+v3.

When we compare this with our original equation, (4), we see that we have a solution y u v= + if

u3+v3= −γ and uv = −β. Thus we have u3+v3 = −γ and u v3 3= −β , 3

which are the sum and product of two quantities, u3 and v3, so these must be the roots of the quadratic equation

w2 +γwβ3 =0 i.e. w=21

RST

− ±γ γ2+4β3

UVW

.

We now know both u3 and v3; let us select

u3 = 12

RST

− +γ γ2+4β3

UVW

and then take the real root (in the case of γ and β real – otherwise any root) for u as U, so

U =3 12

RST

− +γ γ2 +4β3

UVW

. (5) The three roots for u are then

U, ωU , ω2U = − +(1 ω)U, (6)

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where ω = 12

d

− +1 i 3

i

, as described earlier. Since uv= −β, we have the corresponding roots for v:

β

U , β = − = +

ω ω β ω β

U U U

2 (1 ) , β = − ω2 ω β

U U (7) (where we note that ω3= and 11 + +ω ω2 =0). Hence we have the three roots of the cubic

y3+3βy+ =γ 0, with y u v= + from (6) and (7), written as

U

Uβ

, ωU ω β

+ +(1 )U , − +(1 ω)Uω β

U . (8) However, although this is a complete description of the solution, it is only useful if

D=γ2+4β3 is real and non-negative (for then u3 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. 2.1 Case D > 0

With this choice, we have that

U =3 12

RST

− +γ γ2+4β3

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:

UUβ

, β ω β

U U

+

F

+U

HG I

KJ

, − −

F

+

HG I

KJ

U U

ω Uβ ,

which provides the complete solution.

Example 1

Find the roots of the cubic equation x36x− =6 0.

For this equation, we have β= −2 and γ = −6, so γ2 +4β3 =36 32 4 0 = > ; thus we take U = 43 and then β U = −2 34= −32. The three roots are therefore

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4 2

3 +3 , 32+

e

3432

j

ω , 34

e

3432

j

ω ,

where ω = 12

o

− +1 i 3

t

.

2.2 Case D = 0

We now have, from (5), U = −3 γ 2, but this also gives v3 = −γ 2 because the roots of the quadratic for w are repeated. Thus

uv= −β becomes u2 = − >β 0, where we note that 4β3= −γ2 <0 so β< 0 ; hence we have

u v= = ± −β ,

and the sign to be selected must be that which is consistent with y3+3βy+ =γ 0 where y u v= + = ± −2 β .

This clearly requires that u v= = −(sgn )γ β . The three roots are therefore 2u , u

e

ω ω+ 2

j

= −u, u

e

ω ω+ 2

j

= −u,

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 2u and −u (repeated), we must have

(y u+ ) (2 y2u)= y33u y2 2u3 y3+3βy+γ (= 0 )

so β = −u2 and u= −3 γ 2 (both as we obtained above), with γ2 +4β3 =0. Thus with this latter condition applying, the roots are immediately

23 γ 2 and − −3 γ 2 (repeated), selecting the real value of the cube root.

Example 2

Find the roots of the cubic equation 343x384x+16 0= .

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We have β= − 4 49 and γ = 16 343 , so

γ2+4β3=

b

256 117649

g

+ × −4

b

64 117649

g

=0 i.e. D= 0 . Thus we compute 3 γ 2 = −3 8 343= −2 7, and this gives the roots

4 7, 2

7, 2 7.

2.3 Case D < 0

With D=γ2+4β3<0, the function U,

U =3 21

RST

− +γ γ2+4β3

UVW

,

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 µi , then since u3 and v3 are the two roots

w= 12

RST

− ±γ γ2 +4β3

UVW

,

we must have v= −λ µi since one must be the conjugate of the other. (Check:

uv=λ2+µ2 =

{

14

e

γ2γ2 4β3

j }

1 3= −β, which is correct.) The three roots of the cubic are therefore

λ µ λ µ+i + −i , ω λ µ ω

b

+i

g

+ 1

b

λ µi

g

, ω λ µ ω2

b

+i

g

+ 2

b

λ µi

g

i.e. 2λ, ω λ µ ω λ µ

b

+i

g

+ 2

b

i

g

, ω λ µ ω λ µ2

b

+i

g b

+ i

g

or 2λ, ω λ µ ω λ µ

b

+i

g b

+ i

g

, ω λ µ ω λ µ2

b

+i

g

+ 2

b

i

g

,

so we have three real roots (2λ , − −λ µ 3 , − +λ µ 3 ), 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

y3+3βy+ =γ 0 with γ2+4β3 <0,

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so β < 0 ; write β= − (δ δ > 0), and then set y= 2 δY to give

4 3

2 Y3 Y = − γ

δ δ

where γ δ δ2 < , necessarily. Thus we may seek a solution 1 Y = cosθ , for some θ, to produce

4 3 3

2 cos3θ cosθ cos θ γ

= = − δ δ ,

which leads directly to the three real roots for y.

Example 3

Find all the real roots of the cubic equation x36x+ =4 0.

In this case we have β = −2 and γ = 4 , so that γ2+4β3 =16 32 = − <16 0. Set x= 2 2 cosθ , then we obtain

4 3 3 1

2 cos3θ cosθ =cos θ = − ,

and so 3θ =43π,114 π,194 π, ... . The three real roots are therefore

2 2cos

e j

14π =2, 2 2cos

e j

1211π , 2 2cos

e j

1912π .

This trigonometric approach that has been successful in the case

D=γ2 +4β3<0 can be applied more generally (although it cannot give all the solutions in a simple form).

2.4 The trigonometric method We are given

y3+3βy+ =γ 0,

and we introduce y= −2 sgn( )γ βcosθ (where β is not necessarily negative; sgn denotes the ‘signum’ function, giving the sign of γ ) to give

2β β

e

4cos3θ3cosθ

j

= −γ i.e. cos3θ =2(βγ) β .

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Of course, this recovers the result of §2.3 when β< 0 and γ 2(β) − <β 1. If γ2+4β3 =0 (so β< 0 is still the case), then −4β3 = γ i.e. cos3θ =1. Thus θ = 2nπ 3 (n= 0 1 2, , ) and so the three solutions are

2 sgn( )γ β , sgn( )γ β, sgn( )γ β, which is the solution described in §2.2.

This approach fails if D=γ2 +4β3>0, for then we obtain cos3θ >1 which is impossible (for real θ). However if we now introduce y= −2 sgn( )γ βcoshφ (for

β< 0 ), then we obtain

cosh3 ( )

φ 2 γ 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) x3+6x20 0= ; (b) x3+9x− =2 0; (c) x3+18x− =6 0; (d) x32x+ =2 0; (e) x36x+ =2 0.

****************

**********

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3. The solution of the quartic equation

We now consider the general quartic equation

x4+4ax3+6bx2 +4cx d+ =0, (8) which is necessarily quartic by virtue of the term x4. 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.

3.1 Ferrari’s method

The essential idea here is to introduce suitable terms (which cancel identically, of course) into the quartic expression

x4 +4ax3+6bx2+4cx d+ (9) so that this can be written as the difference of two squares. The equation will then take the form X2 Y2 =0 and then X = ± are two quadratic equations each with Y two roots. In order to achieve this, we construct

e

x2+2ax b+ +2y

j b

2 2λx+µ

g

2

=

e

x2+2ax

j b

2 +2b+2y x

ge

2 +2ax

j b

+ +b 2y

g e

2 4λ2 2x +4λµx+µ2

j

, (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

4a2+2(b+2y)4λ2 =6b; 4a b( +2y)4λµ =4c; (b+2y)2 µ2 =d, and we note that the terms x4 and x3 are automatically generated. These three equations yield

λ2 =a2+ −y b; λµ =a b( +2y)c; µ2 =(b+2y)2d which are consistent in the form

b g

λµ 2 =λ µ2 2 if

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a b( +2y)c2 = a2+ −y b b( +2y)2d

or 4y3

e

d+3b24ac y bd

j

+ +2abc c 2a d b2 3=0. (11) This is a cubic (already written in the standard form: no term in y2) which can be solved by the methods described in Chapter 2. 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

a b a b c b c d

+2 =

1

2 2 3 ,

which demonstrates an underlying structure.) The quartic equation can now be written as

x2+2ax b+ +2y+2λx+µ x2+2ax b+ +2y2λxµ =0

e je j

and so the problem is reduced to the elementary exercise of solving two quadratic equations:

x2+2(a+λ)x b+ + +µ 2y=0; x2+2(aλ)x b+ − +µ 2y=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= + a2+ −y b;

(c) determine µ (or λ) from λµ =a b( +2y)c e.g. µ = + + −

a b y c

a y b

( 2 )

2 (and we must do this, rather than from µ2 =(b+2y)2d 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.

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Example 4

Find all the roots of the quartic equation x4+4x− =1 0.

Here we have (see equation (8)): a b= = 0 , c = 1, d = −1, so the resolvent is 4y3+ − =y 1 0; this has a simple root y= 1 2 (we are lucky!). Thus we may use λ= 1 2 and µ= − 2 to produce the two quadratics

x2+ 2x+ −1 2 0= ; x2 2x+ +1 2 0= , with solutions

x= − 1 ±

2

e

1 2 2 1

j

, x= 12

e

1±i 2 2 1+

j

;

these are the four roots of the quartic equation.

3.2 Euler’s method

Again, we start from equation (9):

x4+4ax3+6bx2 +4cx d+ =0

with d ≠ 0; first we remove the x3 term by introducing z= +x a, to give

z4+6αz2+4β γz+ =0 (12) where

α = −b a2, β = +c 2a23ab, γ = −d 4ac+6a b2 3a4.

The significant manoeuvre in this method is to seek a solution of equation (12) in the form

z= ± u± v± w ,

where all sign combinations are allowed (at this stage). This expression, when squared, yields

z2− − − =u v w 2

d id i d id

± u ± v + ± u ± w

i d id

+ ± v ± w

i

,

and when this is squared again:

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z42(u v w z+ + ) 2+ + +(u v w)2

=4

{

uv uw vw+ + +2 u

d id

± v ± w

i d id

+ ±v u ± w

i d id i

+w ± u ± v

}

=4(uv uw vw+ + )+ ±8

d id id

u ± v ± w

id

± u± v± w

i

=4(uv uw vw+ + )+ ±8

d id id

u ± v ± w z

i

.

Thus we have generated a quartic with the z3 term absent:

z42(u v w z+ + ) 2− ±8

d id id

u ± v ± w z

i

+ + +(u v w)24(uv uw vw+ + )=0.

Now this our equation (12) if

3α = − + +(u v w ; ) β= − ±2

d id id

u ± v ± w

i

and γ =(u v w+ + )24(uv uw vw+ + )=9α24(uv uw vw+ + ); i.e. − + +(u v w)=3α; uv uw vw+ + = 14

e

9α2γ

j

; uvw= −41β2.

But these three terms are the coefficients (appropriately) of the cubic equation with roots u, v and w; this cubic is therefore

y3+3αy2+14

e

9α2γ

j

y14β2 =0.

The standard procedure, as we know (Chap. 2), is to remove the term in y2 by writing Y = +y α, which gives

4Y3

e

3α2+γ

j

Y+αγ α 3β2=0. In terms of our original coefficients (see (9)), this becomes

4Y3

e

d+3b24ac Y bd

j

+ +2abc c 2a d b2 3=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= − ±2

d id id

u ± v ± w

i

, so we may write

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z u v

u v

= ± ±

± ±

β 2

d id i

,

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 2

Find all the roots of these quartic equations:

(a) x4 +x35x2 + =2 0; (b) x4 +2x3+x2+8x+ =3 0; (c) x4+5x2 +2x+ =8 0; (d) x411x26x+10 0= ; (e) x4+x3+5x2+5x+12 0= .

****************

**********

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Answers Exercises 1

Throughout, we use ω =21

d

− +1 i 3

i

(a cube root of unity).

(a) 2 1,

d

3

i d

ω+ +1 3

i d

ω, 1 3

i d

ω+ +1 3

i

ω; (b) with u=31+ 28 , v= −3 28 1 , then the three roots are u v u+ , ω+vω ω,u +vω ;

(c) as in (b) with u= 183 , v = − 123 ; (d) approximately − ⋅1 769 0 885, ± ⋅i0 590; (e) approximately − ⋅2 602 2 262 0 340, , .

Exercises 2

(a) − ±1 3,12

d

1± 5

i

; (b) 12

d

− ±3 5

i d

,12 1±i 11

i

; (c) 12

d

1±i 15

i d

,12 − ±1 i 7

i

;

(d) 1± 6,− ±1 3; (e) − ±1 i 2,12

d

1±i 15

i

.

****************

**********

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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 12

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 12

signum 12

Tartaglia 7

trigonometric method 12

References

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19% serve a county. Fourteen per cent of the centers provide service for adjoining states in addition to the states in which they are located; usually these adjoining states have