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Typically, when students first learn about algebraic equations in one variable, the discussion starts with linear equations (i.e. equations of the form ${ax= b}$) and stops at quadratic equations (the general form being ${ax^2+ bx+ c= 0}$) and their solution via the quadratic formula ${x= \frac{-b\pm \sqrt{b^2- 4ac}}{2a}}$. The curious student often wonders why algebraic equations of degree three and more are not discussed. Indeed, many questions come to mind. One may ask if there exist a solution to the general algebraic equation of degree ${n}$

$\displaystyle \begin{array}{rcl} a_nx^n+ a_{n-1}x^{n-1}+ a_{n-2}x^{n-2}+\ldots a_1x+ a_0= 0, \end{array}$

for every positive integer ${n}$ and given coefficients ${a_i}$. If so, how many solutions are there? Furthermore, can each solution be expressed “in radicals”, that is, by a formula that involves only elementary algebraic opertations (addition, subtraction, multiplication, division and extraction of roots)? In this series of posts, we will and answer these questions. The historical notes are taken from the excellent books by Stillwell and Pesic respectively.

Mathematics and its history, by John Stillwell, Springer, 2010.
Abel’s Proof, by Peter Pesic, MIT Press, 2003.

Babylonians (2000BC-600 BC) were probably the first to think of and solve quadratic equations. However, they only considered positive roots since the concept of a negative number did not exist at that time. Brahmagupta (628 AD) provided an explicit formula for the roots of a quadratic which included negative roots as well. Later on, Al-Khwarizmi (825 AD) wrote a treatise on equation-solving, named Al-kitab almukhtasar fi hisab al-jabr w’al-muqabla (The book on restoration and balancing). It is from here that the term Algebra came into being.

The cubic
After the quadratic was solved, not much progress was made in solving equations of higher degree until the sixteenth century when some Italian mathematicians provided the breakthrough. To be precise, del Ferro came up with a solution to the cubic (equation of the form ${x^3+ ax^2+ bx+ c=0}$) in 1520 but he did not publish it (although he did reveal it to a student of his). Later on, this solution was rediscovered by Niccolo Fontana, who was nicknamed Tartaglia (which means “stammerer” in Italian), and the name stuck. Tartaglia fought a scientific duel with del Ferro’s student and beat him badly. This was considered an important problem at that time and the news of this discovery reached Cardano in 1539 and he made several requests to Tartaglia to reveal his solution. The latter resisted the temptation for a while but was eventually persuaded. However, he coded the solution in the form of a poem and asked for an oath of secrecy to never reveal the solution to anyone. However, Cardano was able to inspect del Ferro’s solution (after his death) and he found that the solution found by Tartaglia was already present in del Ferro’s notes. Cardano thus did not feel that he needed to keep his oath and published the result in (his famous book) Ars Magna in 1545. This started a bitter feud as Tartaglia accused Cardano of plagiarism and challenged him to a mathematical duel. Luciodo Ferrari (1545) was a servant in Cardano’s house but he showed such mathematical skills that he soon became a student. It was Ferrari who accepted Tartaglia’s challenge on behalf of Cardano and he won convincingly. As a consequence, Ferrari received a number of big offers, including a professorship at a good university. Soon after, Ferrari was able to solve the general quartic (polynomial equation of degree four).

We now describe the del Ferro-Tartaglia-Cardano solution of the cubic. First, observe that the equation ${x^3+ ax^2+ bx+ c= 0}$ can be reduced to the form ${y^3= py+ q}$ via the substitution ${y= x- a/3}$. Now, let ${y= u+ v}$; substituting this in the previous equation, we get

$\displaystyle \begin{array}{rcl} u^3+ v^3 &=& q\\ 3uv &=& p. \end{array}$

We can eliminate ${v}$ from the second equation so that the first equation becomes
$\displaystyle \begin{array}{rcl} u^6- qu^3+ \left(\frac{p}{3}\right)^3= 0 \end{array}$

which is a cubic in ${u^3}$ and had the roots
$\displaystyle \begin{array}{rcl} \frac{q}{2}\pm\sqrt{\left(\frac{q}{2}\right))^2- \left(\frac{p}{3}\right)^3}. \end{array}$

By symmetry (and the fact that the sum of the roots is ${q}$), the two roots are ${u^3}$ and ${v^3}$ and thus
$\displaystyle \begin{array}{rcl} u^3 &=& \frac{q}{2}+ \sqrt{\left(\frac{q}{2}\right))^2- \left(\frac{p}{3}\right)^3}\\ v^3 &=& \frac{q}{2}- \sqrt{\left(\frac{q}{2}\right))^2- \left(\frac{p}{3}\right)^3}. \end{array}$

Hence, we have
$\displaystyle \begin{array}{rcl} y= u+ v= \sqrt[3]{\frac{q}{2}+\sqrt{\left(\frac{q}{2}\right))^2- \left(\frac{p}{3}\right)^3}}+ \sqrt[3]{\frac{q}{2}-\sqrt{\left(\frac{q}{2}\right))^2- \left(\frac{p}{3}\right)^3}}. \end{array}$

This is the famous del Ferro-Tartaglia-Cardano solution of the cubic.

Complex numbers

As noted in Stillwell, many books on complex numbers mention that complex numbers first came up in the context of quadratic equations but this is false. Neither the Babylonians nor the Indians considered square roots of negative numbers. The greeks too used to think of solutions geometrically and for them such solutions were “impossible.” It was in fact Cardano who considered the case when ${\left(\frac{q}{2}\right))^2- \left(\frac{p}{3}\right)^3< 0}$. In such a case, one is forced to consider imaginary numbers. Note that every cubic equation has at least one real solution (this follows from Bolzano’s theorem). How does one reconcile this with the solution of the cubic? This is exactly the question that Bombelli addressed (1672). He consider the equation ${x^3= 15x+ 4}$, which (by the above method) has a solution

$\displaystyle \begin{array}{rcl} x= \sqrt[3]{2+\sqrt{-11}}+ \sqrt[3]{2- \sqrt{-11}}. \end{array}$

However, it can be verified directly that ${x= 4}$ is a solution as well. Bombelli’s idea was that the terms ${\sqrt[3]{2+\sqrt{-11}}}$ and ${\sqrt[3]{2-\sqrt{-11}}}$be written as ${2+ a\sqrt{-1}}$ and ${2- a\sqrt{-1}}$ and so their sum would be ${4}$ and this was found to be correct. Indeed, you can verify that ${(2+ \sqrt{-1})^3= (2+ \sqrt{-11})}$.

In the next post we will look at Ferrari’s solution to the quartic and then discuss the case of the quintic