# Quadratic, cubic and quartic equations

It is often claimed that the Babylonians (about 400 BC) were the first to solve quadratic equations. This is an over simplification, for the Babylonians had no notion of 'equation'. What they did develop was an algorithmic approach to solving problems which, in our terminology, would give rise to a quadratic equation. The method is essentially one of completing the square. However all Babylonian problems had answers which were positive (more accurately unsigned) quantities since the usual answer was a length.

In about 300 BC Euclid developed a geometrical approach which, although later mathematicians used it to solve quadratic equations, amounted to finding a length which in our notation was the root of a quadratic equation. Euclid had no notion of equation, coefficients etc. but worked with purely geometrical quantities.

Hindu mathematicians took the Babylonian methods further so that Brahmagupta (598-665 AD) gives an, almost modern, method which admits negative quantities. He also used abbreviations for the unknown, usually the initial letter of a colour was used, and sometimes several different unknowns occur in a single problem.

The Arabs did not know about the advances of the Hindus so they had neither negative quantities nor abbreviations for their unknowns. However al-Khwarizmi (c 800) gave a classification of different types of quadratics (although only numerical examples of each). The different types arise since al-Khwarizmi had no zero or negatives. He has six chapters each devoted to a different type of equation, the equations being made up of three types of quantities namely: roots, squares of roots and numbers i.e. $x, x^{2}$ and numbers.
1. Squares equal to roots.
2. Squares equal to numbers.
3. Roots equal to numbers.
4. Squares and roots equal to numbers, e.g. $x^{2} + 10x = 39$.
5. Squares and numbers equal to roots, e.g. $x^{2} + 21 = 10x$.
6. Roots and numbers equal to squares, e.g. $3x + 4 = x^{2}$.
Al-Khwarizmi gives the rule for solving each type of equation, essentially the familiar quadratic formula given for a numerical example in each case, and then a proof for each example which is a geometrical completing the square.

Abraham bar Hiyya Ha-Nasi, often known by the Latin name Savasorda, is famed for his book Liber embadorum published in 1145 which is the first book published in Europe to give the complete solution of the quadratic equation.

A new phase of mathematics began in Italy around 1500. In 1494 the first edition of Summa de arithmetica, geometrica, proportioni et proportionalita , now known as the Suma, appeared. It was written by Luca Pacioli although it is quite hard to find the author's name on the book, Fra Luca appearing in small print but not on the title page. In many ways the book is more a summary of knowledge at the time and makes no major advances. The notation and setting out of calculations is almost modern in style:
                    6.p.R.10
18.m.R.90
____________________________
108.m.R.3240.p.R.3240.m.R.90

hoc est 78.
In our notation
(6 + √10)
(18 - √90) =
(108-√3240 + √3240 - √900)
which is 78.
The last term in the answer 90 is an early misprint and should be 900 but the margin was too narrow so the printer missed out the final 0!

Pacioli does not discuss cubic equations but does discuss quartics. He says that, in our notation, $x^{4} = a + bx^{2}$ can be solved by quadratic methods but $x^{4} + ax^{2} = b$ and $x^{4} + a = bx^{2}$ are impossible at the present state of science.

Scipione dal Ferro (1465-1526) held the Chair of Arithmetic and Geometry at the University of Bologna and certainly must have met Pacioli who lectured at Bologna in 1501-2. Dal Ferro is credited with solving cubic equations algebraically but the picture is somewhat more complicated. The problem was to find the roots by adding, subtracting, multiplying, dividing and taking roots of expressions in the coefficients. We believe that dal Ferro could only solve cubic equation of the form $x^{3} + mx = n$. In fact this is all that is required.
For, given the general cubic $y^{3} - by^{2} + cy - d = 0$, put $y = x + b/3$ to get

$x^{3} + mx = n$ where $m = c - b^{2}/3, n = d - bc/3 + 2b^{3}/27$.
However, without the Hindu's knowledge of negative numbers, dal Ferro would not have been able to use his solution of the one case to solve all cubic equations. Remarkably, dal Ferro solved this cubic equation around 1515 but kept his work a complete secret until just before his death, in 1526, when he revealed his method to his student Antonio Fior.

Fior was a mediocre mathematician and far less good at keeping secrets than dal Ferro. Soon rumours started to circulate in Bologna that the cubic equation had been solved. Nicolo of Brescia, known as Tartaglia meaning 'the stammerer', prompted by the rumours managed to solve equations of the form $x^{3} + mx^{2} = n$ and made no secret of his discovery.

Fior challenged Tartaglia to a public contest: the rules being that each gave the other 30 problems with 40 or 50 days in which to solve them, the winner being the one to solve most but a small prize was also offered for each problem. Tartaglia solved all Fior's problems in the space of 2 hours, for all the problems Fior had set were of the form $x^{3} + mx = n$ as he believed Tartaglia would be unable to solve this type. However only 8 days before the problems were to be collected, Tartaglia had found the general method for all types of cubics.

News of Tartaglia's victory reached Girolamo Cardan in Milan where he was preparing to publish Practica Arithmeticae (1539). Cardan invited Tartaglia to visit him and, after much persuasion, made him divulge the secret of his solution of the cubic equation. This Tartaglia did, having made Cardan promise to keep it secret until Tartaglia had published it himself. Cardan did not keep his promise. In 1545 he published Ars Magna the first Latin treatise on algebra.
Here, in modern notation, is Cardan's solution of $x^{3} + mx = n$.

Notice that $(a - b)^{3} + 3ab(a - b) = a^{3} - b^{3}$
so if $a$ and $b$ satisfy $3ab = m$ and $a^{3} - b^{3} = n$ then $a - b$ is a solution of $x^{3} + mx = n$.
But now $b = m/3a$ so $a^{3} - m^{3}/27a^{3} = n$,
i.e. $a^{6} - na^{3} - m^{3}/27 = 0$.
This is a quadratic equation in $a^{3}$, so solve for $a^{3}$ using the usual formula for a quadratic.
Now $a$ is found by taking cube roots and $b$ can be found in a similar way (or using $b=m/3a$).
Then $x = a - b$ is the solution to the cubic.
Cardan noticed something strange when he applied his formula to certain cubics. When solving $x^{3} = 15x + 4$ he obtained an expression involving √-121. Cardan knew that you could not take the square root of a negative number yet he also knew that $x = 4$ was a solution to the equation. He wrote to Tartaglia on 4 August 1539 in an attempt to clear up the difficulty. Tartaglia certainly did not understand. In Ars Magna Cardan gives a calculation with 'complex numbers' to solve a similar problem but he really did not understand his own calculation which he says is as subtle as it is useless.

After Tartaglia had shown Cardan how to solve cubics, Cardan encouraged his own student, Lodovico Ferrari, to examine quartic equations. Ferrari managed to solve the quartic with perhaps the most elegant of all the methods that were found to solve this type of problem. Cardan published all 20 cases of quartic equations in Ars Magna. Here, again in modern notation, is Ferrari's solution of the case: $x^{4} + px^{2} + qx + r = 0$. First complete the square to obtain
$x^{4} + 2px^{2} + p^{2} = px^{2} - qx - r + p^{2}$
i.e.
$(x^{2} + p)^{2} = px^{2} - qx - r + p^{2}$
Now the clever bit. For any $y$ we have
$(x^{2} + p + y)^{2} = px^{2} - qx - r + p^{2} + 2y(x^{2} + p) + y^{2}$
$= (p + 2y)x^{2} - qx + (p^{2} - r + 2py + y^{2})$ (*)
Now the right hand side is a quadratic in $x$ and we can choose $y$ so that it is a perfect square. This is done by making the discriminant zero, in this case
$(-q)^{2} -4(p + 2y)(p^{2} - r + 2py + y^{2}) = 0$.
Rewrite this last equation as
$(q^{2} - 4p^{3} + 4 pr) + (-16p^{2} + 8r)y - 20 py^{2} - 8y^{3} = 0$
to see that it is a cubic in $y$.

Now we know how to solve cubics, so solve for $y$. With this value of $y$ the right hand side of (*) is a perfect square so, taking the square root of both sides, we obtain a quadratic in $x$. Solve this quadratic and we have the required solution to the quartic equation.

The irreducible case of the cubic, namely the case where Cardan's formula leads to the square root of negative numbers, was studied in detail by Rafael Bombelli in 1572 in his work Algebra.

In the years after Cardan's Ars Magna many mathematicians contributed to the solution of cubic and quartic equations. Viète, Harriot, Tschirnhaus, Euler, Bezout and Descartes all devised methods. Tschirnhaus's methods were extended by the Swedish mathematician E S Bring near the end of the 18th Century.

Thomas Harriot made several contributions. One of the most elementary to us, yet showing a marked improvement in understanding, was the observation that if $x = b, x = c, x = d$ are solutions of a cubic then the cubic is
$(x - b)(x - c)(x - d) = 0$.
Harriot also had a nice method for solving cubics. Consider the cubic
$x^{3} + 3b^{2}x = 2c^{3}$
Put $x = (e^{2} - b^{2})/e$. Then
$e^{6} - 2c^{3}e^{3} = b^{6}$
which is a quadratic in $e^{3}$, and so can be solved for $e^{3}$ to get
$e^{3} = c^{3} +√(b^{6} + c^{6})$.
However
$e^{3}(e^{3} - 2c^{3}) = b^{6}$ so that $b^{6}/e^{3} = -c^{3} +√(b^{6} + c^{6})$.
Now $x = e - b^{2}/e$ and both $e$ and $b^{2}/e$ are cube roots of expressions given above.

Leibniz wrote a letter to Huygens in March 1673. In it he made many contributions to the understanding of cubic equations. Perhaps the most striking is a direct verification of the Cardan-Tartaglia formula. This Leibniz did by reconstructing the cubic from its three roots (as given by the formula) as Harriot claimed in general. Nobody before Leibniz seems to have thought of this direct method of verification. It was the first true algebraic proof of the formula, all previous proofs being geometrical in nature.

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