# Solving The Cubic Equations History Essay

5/12/16 History Reference this

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Cubic equations were first studied In the 11th century by Omar Khayyam, a Persian mathematician and poet. Khayyam discovered that there were 14 different sorts of cubic equation, and remarkably gave geometric justifications for this. However, he did not know about negative numbers, as he believed that all of algebra is a representation of geometry, in which negative numbers do not feature. This means that he would have considered the equations and as of a different type. He also dismissed quartic equations for a similar reason, as he could find no representation of them in regular three-dimensional space. Although Khayyam was able to make progress, he was not able to completely solve the cubic. In 1258, the Mongols sacked Baghdad, destroying all the magnificent libraries, including the House of Wisdom. It is said that the river Tigris ran turned black for months from the ink from the millions of books and manuscripts thrown in the river. The solution of the cubic would have to wait 300 years after the death of Khayyam to be solved by Renaissance Italian mathematicians in the 16th century.

Niccolò Fontana was slashed across the face by King Louis XII’s troops when they invaded his hometown in 1512. This left him with a speech impediment, and it was for this reason that he was known as Tartaglia – “the stammerer.” It was at the time thought that the cubic was insoluble, but Tartaglia was able to use cube roots to help him solve certain types of cubic. However, another Italian mathematician, Antonio Fior, was also claiming to be able to solve the cubic. A contest between the two mathematicians was arranged, where each contestant would set the other 30 cubic equations to solve, and after 40 days a winner would be announced. Just days before the contest was due to start, Tartaglia was able to develop a method that would solve the general cubic equation. Fior only knew how to solve one of the fourteen types of equation, knowledge of which was passed on to him by Scipione del Ferro on his deathbed. In just two hours Tartaglia was able to solve all 30 of Fior’s problems, and his reputation was immediately established.

Girolamo Cardano, another Italian mathematician, heard of Tartaglia’s astonishing successes, and realised that this could only have been done with a formula to solve the general cubic. Cardano spent several years trying to find this formula by himself, but alas in vain. He wrote to Tartaglia asking for him to let him know the secret, on the condition that he would not publish it nor tell anyone else. Eventually, after much persuasion, Tartaglia encrypted the formula in rhyme form, and gave it to Cardano. Cardano was true to his word, except for letting his brightest pupil Lodovico Ferrari in on the secret. Later in the year the young prodigy had managed to adapt Tartaglia’s solution to solve the general quartic equation. Cardano was faced with the dilemma of either breaking his promise to Tartaglia or not allowing his student to receive the praise he deserved. Miraculously, after asking his colleague for advice, he was presented with an old notebook of Scipione del Ferro. Here Cardano found the formula that Tartaglia independently discovered later, and was thus able to justify publishing his Ars Magna in 1545, in which he presented the solutions of both the cubic and quartic, giving Tartaglia due credit.

Needless to say, Tartaglia was furious. He published his account of what had happened, in which he attacked Cardano for what he had done. He then challenged Cardano to a mathematical contest, but Cardano refused to accept. However, it was taken up on his behalf by Ferrari. Letters containing mathematical problems flew between the two and eventually Tartaglia was drawn into open contest with Ferrari. Ferrari revealed his intricate knowledge of Tartaglia’s method, and Tartaglia was forced to flee to Venice, humiliated. Ferrari immediately became famous, and received numerous job offers, but died at the age of 43 of unknown cause. Tartaglia spent the rest of his days in Venice, jobless and penniless.

## Section 2 – Groundwork

## Complex numbers

To facilitate understanding the solution of the cubic equation, one must naturally start off with the preparatory work of considering polynomials of lesser degree. So we shall now consider the solution to the quadratic equation; that is an equation of the form: . Knowledge of the solution to this equation dates to the mathematicians of Babylonia, as early as 2000BC, and the first explicit solution in terms of a formula was given by the Indian mathematician Brahmagupta in 628AD.It is well known that there are two solutions of the quadratic equation [1] , given by the formula . However, this equation raises further fundamental questions, as it leads one to question what happens when the term is negative. The square roots of negative numbers, or imaginary numbers, were unknown to the Greeks, and a comprehensive theory of them was only formulated in the 16th century by Rafael Bombelli, after the solution of the cubic [2] . Nevertheless, imaginary and complex numbers shall be fundamental to our understanding of the cubic equation and its solution, so I shall give a brief introduction of them here.

We start off by defining the imaginary unit as the number with the property that. This clearly is not a real number, as the square of any real number is always greater than or equal to zero. Once we have made this definition, it is obvious how we can begin to use imaginary numbers. For example,. A complex number is a number that is formed of a real part and an imaginary part added together, for example. Complex numbers can be visualised as points on the Argand diagram. This is a two dimensional plane, with orthogonal real and imaginary axes, and complex numbers as Cartesian coordinates, so the complex numberwould be represented by a point reached by going two units along the real axis and one unit along the imaginary axis, both in the positive direction. The general complex number can be written in the form , where a and b are both real. Thus, both imaginary numbers (numbers of the form bi, and real numbers are complex numbers, as can be seen by setting either a or b as zero.

We also define the complex conjugate, , as the number. It can be easily seen that this is just a reflection of in the real axis.

Addition, subtraction and multiplication of complex numbers are obviously defined as follows: , and . Division of complex numbers is less obvious, and is done as follows: consider the number . By multiplying both numerator and denominator by the complex conjugate of the denominator, we obtain: , where the denominator is real, so the division is complete.

We shall now consider two fundamental properties of complex numbers: the modulus and the argument. The modulus of a complex number, is defined as the distance of a complex number from the origin in the Argand diagram, which is clearly, so from our example the modulus ofis. The argument is the angle, measured in radians such that, between the line joining the complex number to the origin and the positive real axis. So for our example above, the argument of would be such that , so to 3sf.

Complex numbers can be uniquely defined by a modulus and argument, much like polar coordinates, where a point is defined by distance from the origin and angle subtended with the positive x-axis. This leads to the polar form of a complex number. If , and , then we can see that and , so , the polar form of a complex number.

A result that is of vital importance when considering complex numbers, and which it shall be necessary to utilise later in our solution of the cubic, is known as De Moivre’s Theorem. This states that ifare two complex numbers with polar forms and , then the product . This is equivalent to saying that when we multiply two complex numbers together, we multiply the moduli and add the arguments, as can be seen by comparing with the polar form of a complex number, given above. The proof of this is as follows:

## 3.

I shall now introduce another notation for complex numbers, brought about by Leonhard Euler. This is the exponential form:, and is a direct consequence of Euler’s formula, described by Richard Feynman as ‘one of the most remarkable, almost astounding, formulas in all of mathematics.’ Using this notation, De Moivre’s Theorem becomes a triviality, as clearly. This notation is especially useful when considering the cube roots of unity. First we must note that .Note also that using the definition of the argument of a complex number, adding or to the argument of a complex number will have no effect. From this it follows that the three cube roots of unity are the numbers that satisfy, in other words.

## Polynomials

How do I necessarily know that there will be three, and only three, cube roots of 1? This is a direct consequence of the Fundamental Theorem of Algebra, whose sheer importance is stressed by its name. In one form this theorem states that any non-zero polynomial in one variable with complex coefficients has as many complex roots as its degree [4] , including roots that have been repeated. Thus, the polynomial has exactly three roots, which correspond to the values of for which. And, more generally, any cubic equation of the form will have exactly three complex roots, including possible repeats [5] .

Next I am going to consider real polynomial equations, that is polynomial equations with strictly real-valued coefficients. I wish to prove a result about the roots of such an equation, but to do so I shall first need to prove a complimentary result, that for all complex numbers z, , where * represents the complex conjugate:

Consider two complex numbers: and. First I shall show that. Let us consider the left hand side of the equation and expand it:. And now the right hand side:. This shows that the initial equation is true. Setting w as z, we obtain the result: .And if we use the two complex numbers and, we obtain the result:. It is not hard to see that this easily extends to the general case:.

Now let us consider the real polynomial equation. Let the complex number be a root of this equation, that is. Now we shall show that the complex conjugate is also a root. Note that for all the coefficients of the polynomial, as they are real:

## .

This shows us that if a complex number is a root of a real polynomial, then its conjugate is also a root, and this result will be useful later when dealing with real cubics.

## Section 3 – A First Glance at the Cubic

If the coefficients of a cubic are real, then we can use the Fundamental Theorem of Algebra and the result we proved in the last section to deduce four possible scenarios for the roots:

There are three distinct real roots,

There is one distinct and two repeated real roots,

There are three repeated roots,

There is one real root, and two complex conjugate non-real roots.

Note that there must always be at least one real root of a real cubic, as if , then if a is positive, and the other way around if a is negative. Therefore, using continuity, the graph of y must cross the x-axis at some point, so there is at least one real root.

If we differentiate the general cubic function given above:, we will be able to determine when the curve will have turning points, which will help in our understanding of the roots. So, differentiating, we get . At a turning point, the value of the derivative will be zero, so we wish to solve in general the equation. By the quadratic formula, this has solutions, which leaves us with three cases to distinguish between:

If , then there are no turning points, so there is only one real root.

If , then there is only one turning point, and so there is one real root which may be a triply repeated real root if the turning point is on the x-axis (see diagrams)

If , then there are two turning points, which we shall denote 1() and 2. We shall now examine this third case further.

Clearly the x-coordinates of the turning points are: ,. Thus, if a is positive, is to the left of, and vice versa.

One could calculate the second derivative and plug in the values of andto determine for which values of x 1 and 2 are maxima and minima, but this would be a messy and long-winded approach. If instead we assume without loss of generality that is to the left of, we can consider and . This constitutes a much simpler approach. We need not trouble ourselves with calculating what they are; this would be very messy [6] .

Now there are 10 different cases to consider:

When, , we have a positive cubic with one real root.

When, , we have a negative cubic with one real root.

When, , we have a positive cubic with one distinct and two repeated roots.

When, , we have a negative cubic with one distinct and two repeated roots.

When, , we have a positive cubic with three distinct roots.

When, , we have a negative cubic with three distinct roots.

When, , we have a positive cubic with one distinct and two repeated roots.

When, , we have a negative cubic with one distinct and two repeated roots.

When, , we have a positive cubic with one real root.

When, , we have a negative cubic with one real root.

These 10 cases are illustrated in the diagram below:

## Section 4 – The Solution of the Cubic

We start off by ensuring that our cubic equation is monic, that is the leading coefficient [7] is 1. This is easily done by dividing the equation by the leading coefficient, leaving us with the general cubic equation: (1). The first step to solving the cubic is to eliminate the term, which makes the calculations simpler, as the Italian mathematicians of the 16th century found out. This is done by letting . From this, we can deduce that: (2), so (3), for some constants A and B. We shall write this as (4), for some constants h and k which can easily be worked out.

We have succeeded in eliminating the term, and now have a cubic equation in y. The solution from here is known as Cardano’s solution [8] . We write , so that . Therefore the equation (5) has as a root. Comparing this with (4) leads us to set and . From this we can deduce that, and hence that . Multiplying this out, we get (6), which is just a quadratic in . As (5) is symmetric in and , equation (6) is equally valid if we replace by. As and are the roots of (6), we can take either root of the quadratic as being equal to without any loss in generality. Thus, we take ,. Now we recall that is a root of (5), which gives us the following formula for the roots of the cubic: .

Whilst it may at first appear that this formula can take nine different values, as there are three different cube roots of and , this is not the case, each value of uniquely determines the value of by given above.

It is here that our work from Section 2 becomes vital. For there are three cube roots of , and to express these we require complex numbers. We saw above that the cube roots of 1 are , and so likewise the cube roots of are: , which can be verified by cubing them. We shall simply call , and thus the cube roots are: . Therefore we know that the solutions to the cubic equation (5) are , , , which can be written as ,, . The final step is to convert these solutions of (5) into solutions of (1) by the simple transformation . We have now solved the general cubic equation.

## Nature of the roots of the cubic

I would like to now focus on , which is found in the formula for the roots of the cubic. I must remind the reader that we are only considering cubics whose coefficients are real. If , then the square root is real and we have no problem, we will end up with one real root, , and two complex roots. If , then our roots become ,, and thus the equation has a repeated root. If we call , then , where is this repeated root. Differentiating, we get: . Thus, the equation also has as a root. Therefore, is a root of the equation, which can very easily be solved, eliminating need for the method outlined above. It can be shown that in this case there are three real roots, two of which are the same. However, in the case when , something rather different occurs:

## Casus Irreducibilis

When , its square root is imaginary, which results in and being complex. However, it is always the case that such cubics have three distinct real roots, but that these cannot be expressed in terms of radicals without using complex expressions at some point. This is known as casus irreducibilis. We could calculate the cube roots of complex numbers but this requires very messy calculations. However, although we cannot evaluate our answers exactly, we can numerically approximate the real solutions using a trigonometric method invented by Vieta:

We are considering the equation . We know , so as k is real, then h must be negative. We set . This transforms our original cubic into the form , from which by the trigonometric identity . This enables us to numerically find three values of , so we can find the three roots of the equation. Notice again that all three roots of this equation will always be real.

To illustrate this problem with an example, let us consider the cubic equation . We have and,thus . Using Cardano’s method, a solution to the equation is . However, with a little thought our original cubic equation can be factorised as , and so has solutions . Which then, of our solutions is equal to the rather ugly expression two lines up? A fair amount of work reveals it to actually be equal to 2, yet the expression involves complex numbers. The Italian mathematicians of the 16th century came up against this problem, and to a large extent this motivated the study of imaginary and complex numbers. Using Vieta’s method, we end up with , from which a solution is , leading to However, we cannot be sure that this root is exactly equal to 2 in this case, as Vieta’s method is purely numerical.

## Section 5 – Post-mortem and Extensions of the Problem

## Alternate numerical method using calculus

This is a much faster way of numerically approximating the roots of any real cubic. Our first step is to numerically approximate a real root of the cubic [9] . To do this we shall use the Newton-Raphson method for numerically approximating the roots of an equation. I shall give a brief outline of this method.

Graphically, this method can be though of as constructing a tangent to a curve, finding the point where this line crosses the x-axis, moving vertically up from that point back to the graph, and repeating the process. We shall develop this into a formula for the numerical approximation. Let us start off by calling the equation of the cubic [10] . The point on the curve that corresponds to our starting x-value, , is . Now, let us take a tangent from here, and let the point where this tangent crosses the x-axis be called , our next approximation to the root of the cubic. The gradient of this tangent is the value of the derivative of at , also known as . The gradient can also be calculated by , therefore . Setting these two as equal to each other gives, , and rearranging, we end up with . We can then define the iteration process , which enables us to calculate better and better approximations from our initial starting value.

As long as our initial approximation is fairly accurate [11] , then the Newton-Raphson method converges extremely rapidly on the root, enabling us to calculate it to any level of accuracy we wish. Having used the Newton-Raphson method to numerically find one of the roots of the cubic, we can then factorise the cubic to leave us with a quadratic, which we can solve using the quadratic equation to give either two real roots, one repeated real root or two complex conjugate roots. If is a root of , then , so we can solve via the quadratic equation to find the other two roots. I shall now illustrate this method using the cubic equation . We shall use the initial approximation . , so we can calculate . Repeating this process, we have:

In just 5 steps we have obtained ten significant figures of accuracy, and we could easily continue this process to obtain any level of accuracy we desired.

We now know that is a root of to 3sf. Using this information, we can factorise , using one extra significant figure that we require for our end result. Finally, we solve by the quadratic formula to obtain the other two roots of the equation, to 3sf.

## Higher Degree Polynomials

We have shown in Section 3 the method by which the general cubic can be converted to a depressed cubic of the form by a simple substitution. Similarly, Ferrari discovered in 1540 that by a suitable substitution the general quartic [12] equation can be converted to a depressed quartic of the form , which has no term [13] . Having depressed the quartic to this form, Ferrari managed through a series of very clever algebraic manipulations and adding identities to the depressed quartic to eliminate a root and convert it to a cubic equation. This is then solved as above, by again converting to its depressed form, then using Cardano’s method.

Both the cubic and the quartic can be solved by radicals, that is an expression involving a sequence of arithmetic operations as well as the extraction of roots (square, cube, fourthâ€¦) which, when applied to the coefficients of the original equation, returns the roots of that equation. The cubic and the quartic were both solved in the same century, and it was only natural after that to turn one’s gaze to the quintic, the general fifth degree polynomial. However, several centuries passed without anyone being able to solve this equation in terms of radical expressions, and for good reason – no such formula exists! This was proved in 1824 by Abel and Ruffini, and is now known Abel-Ruffini theorem. This does not mean that all quintics are insoluble: consider for example , which can be factorised as , and thus has roots at . However there are some quintics equation for which the roots can not be expressed in terms of radicals, an example is . Five complex solutions to all quintics equations still exist, and these can be practically found by numerical methods. The technique mentioned above would also work; we can find a root by the Newton-Raphson iteration and then factorise to leave a quartic which can be solved.

After the publication of the Abel-Ruffini theorem, Évariste Galois, a French prodigy in his teens, formulated a method that would enable him to determine whether any polynomial equation was soluble by radicals. Galois proved that not only does the general quintic not have a radical formula for its roots, but that there is also no formula for all general polynomials of a higher degree. His work would later be formalised into group theory and Galois theory. Galois lived until just 20 years old, when his died after being shot in a duel over a mysterious woman. Nevertheless, his ideas, formulated in the 19th century, are central to a very substantial portion of modern mathematics. I will now outline the basic ideas behind his proof of the insolubility of the quintic equation.

## The Group Axioms

We define a group as a set G and an operation that satisfy four conditions:

For any elements in G, is also in G.

is associative, that is for any in G.

There is an identity element in G such that for all in G.

For any element in G there exists in G the inverse of , , such that .

As an example, let G be the set of rotational symmetries of a triangle, and the composition of symmetries. Thus , where represents a rotation of clockwise, and represents a rotation of clockwise. Clearly the identity element leaves the triangle where it is. Also, the composition of any two rotations of the triangle is identical to another rotation, for example . Finally, each element has an inverse; notice that , so the inverse of isand vice versa, and the identity is its own inverse. Thus the set and the operation we have chosen truly do form a group.

Galois then considered the group of permutations of the roots of an equation that maintain all true equations with rational roots in the original order of the roots, aptly called the Galois group of an equation So for example, consider and . These are indistinguishable using equations with rational coefficients, as , so and , so . Thus, if and were two of the roots of an equation, we would be able to permute them without affecting these true equations.

Galois discovered that by examining the Galois group of an equation you can decide whether or not the equation is soluble by radicals. This is done by examining whether the Galois group contains a normal subgroup [14] . If the number of elements of the original group divided by the number of elements of the subgroup is prime, and if the subgroup contains another subgroup, and if the ratio of the numbers of elements is again prime, and so on until we reach a subgroup containing only one element of the original group, them the equation is soluble.

Conclusionâ€¦.?

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