b

–

4a

B

–

4

e

–

a

d

–

a

c

–

a

b

–

a

A

quartic equation 431

2

that all natural numbers of the form 2n+ 1 are odd is

written:

n, 2n+ 1 is odd

that there exists a natural number nsatisfying n×n= n

+ ncan be written:

∃n: n×n= n+ n

(a colon is usually read as “such that”), and that there is

no natural number nsatisfying n×n×n= n+ n+ nas:

¬

(∃n: n×n×n= n+ n+ n)

See also

ARGUMENT

;

NEGATION

.

quartic equation (biquadratic equation) Any degree-

four

POLYNOMIAL

equation of the form ax4+ bx3+ cx2

+ dx + e= 0 with a≠0 is called a quartic equation.

Italian scholar L

UDOVICO

F

ERRARI

(1522–65),

assistant to G

IROLAMO

C

ARDANO

(1501–76), was the

first to find a general arithmetic formula that would

solve for xin any quartic equation. His method was

published by Cardano in the 1545 epic work Ars

magna (The great art). French mathematician R

ENÉ

D

ESCARTES

(1596–1650) also found a method of solu-

tion, which we briefly outline here.

By dividing through by the leading coefficient

a, we can assume that we are working with a

quartic of the form:

x4+ Bx3+ Cx2+ Dx + E= 0

for numbers B= , C= , D= , and E= .

Substituting x= y– simplifies the equation

further to one without a cubic term:

y4+ py2+ qy + r= 0

This form of the quartic is called the reduced

quartic, and any solution yto this equation

corresponds to a solution x= y– of the

original equation.

Assume that the reduced quartic can be fac-

tored as follows, for some appropriate choice

of number λ, m, and n:

y4+ py2+ qy + r= (y2+ λy+ m)(y2– λy+ n)

EXPANDING BRACKETS

and

EQUATING COEFFI

-

CIENTS

consequently yields the equations:

Summing the first two equations gives n=

; subtracting them yields m=

; and substituting into the third

equation yields, after some algebraic work, a

cubic equation solely in terms of λ2:

(λ2)3+ 2p(λ2)2+ (p2– 4r)(λ2) – q2= 0

C

ARDANO

’

S FORMULA

can now be used to

solve for λ2, and hence for λ, m, and n. Thus

solutions to the quartic equation:

y4+ py2+ qy + r= (y2+ λy+ m)(y2– λy+ n) = 0

can now be found by solving y2+ λy+ m= 0

and y2– λy+ n= 0 using the

QUADRATIC

formula.

This method, in principle, is straightforward, but

very difficult to carry out in practice. It does show,

however, that, if one has the patience, one can indeed

write down a formula for the solution of a quartic

equation ax4+ bx3+ cx2+ dx + e= 0 using only the

numbers a, b, c, d, and e, and their roots.

During the 1700s there was great eagerness to find

a similar formula for the solution to the quintic

(degree-five equation). L

EONHARD

E

ULER

(1707–83)

attempted to find such a formula, but failed. He sus-

pected that the task might be impossible.

In a series of papers published between the years

1803 and 1813, Italian mathematician Paolo Ruffini

(1765–1822) developed a number of algebraic results

that strongly suggested that there can be no procedure

for solving a general fifth- or higher-degree equation in

a finite number of algebraic steps. This claim was

indeed proved correct a few years later by Norwegian