Girolamo Cardano 1500-1576

Cardano’s autobiography, ‘The Book ofmy Life’, led William Dunham todescribe him as ‘perhaps the mostbizarre character in the whole history ofmathematics’. After looking at some ofthe strange episodes in his life we’llfocus on the algebra in his famousbook ‘Ars Magna’ (or ‘The Great Art’),including working through the solutionof the cubic and quartic equations.

Suitable for all teachers of GCSE and Alevel mathematics; fluency withalgebraic manipulation is all that isexpected.

‘[Before Cardano] there is absolutely notrace of scientific self-study. This fact helpsone to realise what a wholly fresh ideacame to the Italian physician when he set towork examining himself ‘as if he were anew species of animal which he neverexpected to see again’……The fact remains that here, in 1575, sitsthe first psychologist, minutely examiningthe only case at hand, a case which,fortunately for us, presents the most salient,individual and often abnormal features. It isnot only that a scientist of the first order isexamining this brain; but the brain itself isof the first order,...’

Anna Robeson Burr, ‘The Autobiography’

“Cardano was a great man with all his faults;without them he would have been incomparable.”

Gottfried Wilhelm Leibniz

3

3

3

3 2

x mx n

x mx n

x n mx

x lx mx n

Scipione del Ferro (1465–1526) knew how to solve equations ofthe form x3+mx=n and on his deathbed told

Antonio Fior (1506 - ?) who challenged

Niccolo “Tartaglia” Fontana (1499-1557) who could already solveequations of the form x3+mx2=n . He discovered the solution tox3+mx=n on February 13th 1535 and so triumphed in thechallenge. On March 25th 1539 he told, under oath of secrecy,

Girolamo Cardano (1501 – 1576) who, with his student

Ludovico Ferrari (1522 – 1565) read del Ferro’s work in 1543and published Ars Magna in 1545.

2 3 2 33 3 3

2 4 27 2 4 27

n n m n n mx mx n x

Del Ferro’s solution of the cubic

3 215 81 175 0x x x

2 3 2 33 3 3

2 4 27 2 4 27

n n m n n mx mx n x

The quartic

The quartic

The quartic – Ferrari (1540)

Write this in the form

2

22 02

ax x p qx r

4 3 2 0x ax bx cx d

4 3 24 36 16 128 0x x x x

The quartic – Euler (1750)

4 2

22

3 2

0 has root

Show 4 8

, ,

Relate to cubic 0

i.e. to 0

x ax bx c x p q r

x f g hx

f p q r g pq qr rp h pqr

z p z q z r

z fz gz h

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MEI Conference 2016

Girolamo Cardano1501-1576

Bernard Murphybernard.murphy@mei.org.uk

Main references

‘The Book of My Life.’ Cardano’s Autobiography. First English translation 1930.http://djm.cc/library/cardan-book-of-my-life-1930.pdf

Ars Magna (1545). English translation in 1968, with modern algebraic notation.https://albeniz-matematicas-acaro.wikispaces.com/file/view/Richard+Witmer+(1968)+Girolamo+Cardano+ARS+Magna+or+the+Rules+of+Algebra.pdf

Del Ferro’s method for solving the general cubic

1. Divide every term of the cubic 3 2 0ax bx cx d by a, the coefficient of 3x .

2. Substitute3

bx y

a to give a depressed cubic (i.e. one with no quadratic term):

3y my n

3. Compare the identity 3 3 33t u tu t u t u with the depressed cubic.

4. Solve the simultaneous equations 3 33 ,tu m t u n to find t and u in terms of mand n.

5. A root of the depressed cubic is y t u and so a root of the original cubic is

3 3

b bx y t u

a a

2 3 2 33 3 3

2 4 27 2 4 27

n n m n n my my n y

Exercise3 2 3 3

3 2 3 3

3 2 3 2

1. 15 81 175 0 2. 15 4 3. 9 26

4. 3 3 4 0 5. 6 4 6. 6 40

7. 6 11 6 0 8. 6 13 12 0

x x x x x x x

x x x x x x x

x x x x x x

Example. Find the roots of 3 215 81 175 0x x x .

First use the substitution 5x y to get a depressed cubic:

3 25 15 5 81 5 175 0y y y which simplifies to 3 6 20y y .

Using the formula with 6, 20m n :

2 3 2 3

3 320 20 6 20 20 6

2 4 27 2 4 27y

which simplifies to 3 3 3 310 108 10 108 10 6 3 10 6 3y

But 31 3 10 6 3 and 31 3 10 6 3 .

Therefore 1 3 1 3 2y , a root of 3 6 20y y .

It follows that 5 2 5 7x y is a root of 3 215 81 175 0x x x .

To find the other two roots, divide this polynomial by 7 :x

23 2 215 81 175 7 8 25 7 4 9x x x x x x x x

The roots of the equation 3 215 81 175 0x x x are 7, 4 3 , 4 3x i i .

-----------------------------

You may prefer not to use the formula but go back to basics and compare the depressed

cubic 3 6 20y y with the identity 3 3 33t u tu t u t u .

Then 3 6tu and 3 3 20t u . Substituting2

ut in 3 3 20t u leads to the quadratic in

3t : 23 320 8 0t t which, on completing the square, gives 23 10 108t and so

3 10 108 10 6 3t . Substituting this in 3 3 20t u gives 3 10 6 3u .

Therefore 3 310 6 3 10 6 3t u which simplifies to

1 3 1 3 2t u and this is a root of the equation 3 6 20y y .

Ferrari’s method for solving the quartic

1. Divide throughout by the coefficient of 4x to give 4 3 2 0.x ax bx cx d

2. Compare the coefficients of this quartic with those of 2

22 0.2

ax x p qx r

3. By eliminating q and r from the resulting simultaneous equations, form a cubic in p.

4. Use Cardano’s method (or other) to find p and then, by substitution, q and r.

5. Solve the two quadratics 2 .2

ax x p qx r

Example Find the roots of 4 3 24 36 16 128 0x x x x

Compare the quartic with 2 22 2 0x x p qx r :

Coefficients of 2 2 2: 2 4 36 2 40x p q q p 1Coefficients of : 4 2 16 2 8x p qr qr p 2Constant term: 2 2 2 2128 128p r r p 3

Now express 2 2q r in two ways: 2 22 8 2 40 128p p p which simplifies to3 218 144 2592 0p p p .

Cardano’s method is an option but this cubic factorises: 18 12 12 0p p p .

Using 12p (interesting later to try -12 or 18) in 1 , 2 and 3 gives

8, 4q r and 32qr , the latter showing that q and r must have opposite signs.

Substituting these back into 2 22 2 0x x p qx r gives

2 22 2 12 8 4x x x and so 2 2 12 8 4x x x , i.e. two quadratics to solve.

2 22 12 8 4 10 16 0 2 8 0x x x x x x x

and 2 22 12 8 4 6 8 0 2 4 0x x x x x x x

Therefore the four roots of 4 3 24 36 16 128 0x x x x are 2,8, 2, 4.x

Exercise

4 3 2 4 3 2

4 4 3 2

4 2 4 3

4 2

1. 8 14 4 8 0 2. 10 35 50 24 0

3. 16 12 0 4. 6 12 12 4 0

5. 25 60 36 0 6. 4 8 32 0

7. 28 48 0

x x x x x x x x

x x x x x x

x x x x x x

x x x