Poly by emonarrez

Polynomials

Nkechi Agwu Borough of Manhattan Comm. Coll. New York, New York

Patricia Frey Buffalo Academy for the Arts Buffalo, New York

Tarhonda Greer Jesse Jones Senior High School Houston, Texas

Gwendolyn Taylor Jack Yates Senior High School Houston, Texas

Carole Gould Franklin High School Franklin, Tennessee

JoAnn Perkins Western Albemarle High School Crozet, Virginia

Thanh Le Lynwood High School Glendale, California

Victor J. Katz University of the District of Columbia Washington, DC

Historical Modules Project Sponsored by the Mathematical Association of America Funded by the National Science Foundation under grants ESI-9730329 and ESI-0120923

ÂŠ2004 by the Mathematical Association of America

Table of Contents Introduction Suggested Courses for Using Module Activities

3 5

Historical Introduction Babylonian Mathematics Chinese Mathematics Indian Mathematics Greek Mathematics Islamic Mathematics Western Mathematics Biographies

6 7 7 8 9 10 11 12

Quadratic Equations Finding Square Roots Deriving the Quadratic Formula Problems Involving Quadratic Equations Descartes’s Method for Solving Quadratic Equations Geometrically

19 19 33 38 47

Cubic Equations Finding Cube Roots Solving Cubic Equations via Conic Sections Solving Cubic Equations Algebraically Introduction to Complex Numbers Solving Cubic Equations Using Trigonometry

50 50 57 62 79 88

Higher Degree Polynomial Equations Solving Quartic Equations Newton’s Method for Solving Equations The Relationship Between Roots and Coefficients of a Polynomial Finding Maxima and Minima of Polynomials

92 92 95 98 108

Bibliography

113

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Introduction The Principles and Standards for School Mathematics, released by the National Council of Teachers of Mathematics to begin the 21st century, advise the reader that “the secondary school mathematics program must be both broad and deep.” (NCTM 2000, 287) The study of polynomials brings a way to analyze mathematical form and structure because of the shared properties with other functions. This results in an awareness of the connections over time as we add the historical perspective to the study of polynomials. Many classroom teachers have never combined a historical information base with their instruction. When provided such a basis, students can, in truth, “build on prior knowledge,” and attain more depth to their understanding of the processes. Problems can be offered that provide real world application and bring more meaning to the student. This adds to the development of problem-solving skills and “weaves together [other] content strands.” (NCTM 2000, 289) As the Standards so clearly state: We owe our children no less than a high degree of quantitative literacy and mathematical knowledge…” (NCTM 2000, 289) This module deals primarily with the solution of polynomial equations of degrees two and three, with some attention to polynomials of higher degree. It provides a historical background for each of these methods and helps students understand the various types of solutions which may be useful for a given equation. The material here will help the user to decide “whether a problem calls for a rough estimate, an approximation, or an exact [response].” A thorough study of the activities in this module will provide students with an introduction to the more general study of functions and functional equations.

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How to Use This Module This module contains numerous activities designed to teach methods of solving polynomial equations via the historical ideas which led to these methods. Some of the activities can be used to replace corresponding sections of a course text in Elementary or Intermediate Algebra or Precalculus. Other activities are for enrichment and can be used with individual students or small groups to deepen their understanding of the material. In particular, few secondary courses deal explicitly with the general solution of cubic or quartic equations. Yet the methods for solving these are not inherently difficult. They are just a bit “messier” than the corresponding methods for solving quadratic equations. And the solution formula for cubic equations helps students understand the necessity for the introduction of complex numbers. Thus, it is very useful for students to learn how to solve cubic equations, at some level of detail. This module provides numerous opportunities for students to work through the various methods of solution and gain an understanding of equation solving deeper than one would get by only solving linear and quadratic equations. In general, one important idea that needs stressing with students is the meaning of the term “solution” of an equation. As a simple example, teachers could discuss the equation x2 = 2. What does it mean to have a “solution”? Is x = 2 a solution. Is x = 1.4142… a solution? Is the diagonal of a square of side 1 a solution? In some sense, the answer to each of these questions is “yes.” But it is up to the teacher to inculcate in the students the notion that different types of solutions are appropriate in different situations. And whether one of these solutions is acceptable may well depend on the nature of the problem from which the equation came. The historical introduction at the beginning of the module contains general background information for the teacher as well as biographical sketches of some of the historical figures important in the development of methods of solution of polynomial equations. This introduction can be shared with the students. References for further investigation are noted there. The chart on the next page shows the possible courses in which each activity could be used.

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Suggested Courses for Using Module Activities

Activity

Prealgebra

Algebra 1

Algebra 2

Finding Square Roots Deriving the Quadratic Formula Problems Involving Quadratic Equations Descartes’s Method for Solving Quadratic Equations Finding Cube Roots Solving Cubic Equations via Conic Sections Solving Cubic Equations Algebraically Introduction to Complex Numbers Solving Cubic Equations Using Trigonometry Solving Quartic Equations Newton’s Method for Solving Equations Relationship Between Roots and Coefficients of a Polynomial Finding Maxima and Minima of Polynomials

X X X

Precalculus/ Trigonometry

X X X

X X

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Historical Introduction Origin of the Term Algebra Algebra is a Latin variant of the Arabic word al-jabr (sometimes written al-jebr). Al-jabr means “the reunion of broken parts.” This terminology gained widespread use through the book, Al-kitab al-muhtasar fi hisab al-jabr w’al’muqabalah (The Condensed Book on the Calculations of al-jabr and al-muqabala — the science of restoring missing terms and equating like terms), written by Abu-Abdullah Mohammed ibn-Musa Al-Khwarizmi (Mohammed, father of Abdullah and son of Moses, from Khwarizm, now in Uzbekistan), in Baghdad, Iraq, around 825 CE. Algebra, which also had roots thousands of years earlier in Babylon and Egypt and was known in India in the sixth century, was transmitted by the Arabs to Europe via Italy and Spain. Algebra was applied to various aspects of human life by all of these cultures. But it did have a theoretical basis, taken from geometry, which was emphasized by Islamic writers after alKhwarizmi, who were familiar with the geometric algebra of the Greeks.

Development of Algebraic Notation The development of algebraic notation progressed through three stages: rhetorical (verbal), syncopated (use of abbreviations), and symbolic. The mathematics of the ancient Babylonians, Egyptians and Chinese was purely rhetorical. Problems and their solutions were written out completely in words. With time, printing and the repeated use of certain words led algebra to enter the syncopated stage, where abbreviations were used for some common words and operations. Later, these abbreviations evolved into symbols. In this third, symbolic stage, notation went through many modifications and changes, with each local group of mathematicians having their own system of symbolization, until it became fairly stable by the time of Isaac Newton (c. 1700). It is interesting to note that even today there is some lack of uniformity in the use of symbols. In 150 CE Diophantus of Alexandria used a letter to stand for the unknown. And there was some symbolization in medieval Indian algebra. But the algebra that reached Europe in the twelfth century was the rhetorical algebra of al-Khwarizmi and other Islamic mathematicians. Thus it was in Europe that our symbolic notation developed. During the Renaissance, Italian mathematicians began to use abbreviations for various common mathematical terms, and also to use such symbols as those for addition, subtraction, and equals. Symbols were also developed for the unknown in an equation as well as its various powers In 1590, the Frenchman François Viète expanded this idea, by using letters to represent not only unknowns but also the constants in an equation. Thus, he used vowels to represent the quantities in algebra that was assumed to be unknown and consonants to represent numbers assumed to be known or given. Viète’s algebra is fundamentally syncopated, however, rather than symbolic, because he used abbreviations to represent such concepts as multiplication, powers, and even “is equal to.” Thomas Harriot, in the 1620s, turned Viète’s ideas into a fully symbolic algebra. Unfortunately, Harriot never published his ideas; they only had a limited circulation in manuscript in England. So it was René Descartes’s reformation of Viète’s symbolism, which appeared in print in 1637,

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which became the notation now commonly used. Descartes introduced the use of letters at the end of the alphabet to represent unknown quantities, and those at the beginning of the alphabet to represent known quantities, the system still in general use today. In what follows, we give brief accounts of some of the civilizations and people involved in the development of methods of solving polynomial equations.

Babylonian (Mesopotamian) Mathematics (3500 BCE - 600 BCE) The ancient civilization of Mesopotamia is one of the oldest civilizations for which we can trace the development of mathematical ideas. It is popularly known as the Babylonian civilization after the city whose ruler (Hammurapi) conquered this area around 1700 BCE It dates as far back as 3500 BCE in the Tigris and Euphrates river valleys. It continued almost till the dawn of Christianity, even after Babylon fell to Cyrus of Persia in 538 BCE. The principal sources of information on Babylonian mathematics are the clay tablets with writings in cuneiform script, taken from the sites of ancient Nippur and Uruk, among many others. Many of these tablets are now housed in university libraries around the world, including Columbia, Pennsylvania and Yale. These tablets show that the Babylonians had a positional numeration system and that they could work with sexagesimal fractions. They indicate that the Babylonians were skillful in developing algorithmic procedures such as a process for calculating square root, that they had developed tables analogous to our modern day exponential and logarithmic tables, and that they were familiar with the concept of linear interpolation. They also show that the Babylonian algebraic skills were sophisticated to the point of solving linear, quadratic, and even some higher order equations. See [Neugebauer] and [van der Waerden].

Chinese Mathematics (1600 BCE - 220 CE) The earliest evidence of the mathematical traditions of the ancient Chinese civilization is provided by the “Oracle bones,” excavated at Anyang near the Huang River. These bones, which date back to 1600 BCE, the period of the Shang Dynasty, are a principal source of knowledge of the Chinese numeration system. At the beginning of the first millennium BCE, the Shang Dynasty was replaced by the Zhou Dynasty which later dissolved into several feudal states. This period saw much intellectual and technological growth, especially in the sixth century BCE The feudal period ended in 221 BCE with the unification of China under Emperor Qin Shi Huangdi. His leadership resulted in the transformation of China into a highly centralized bureaucratic government which enforced legal codes, levied taxes evenly, and standardized weights, measures, money and writing. The Qin Dynasty was overthrown and replaced by the Han Dynasty with the death of Qin in 210 BCE. The Han Dynasty lasted for 400 years, completing the development of a highly skilled and educated civil service workforce that had texts to assist in performing their duties. One of these texts is the Jiuzhang Suanshu (Nine Chapters on the Mathematical Art), our principal source of information on ancient Chinese mathematics. See [Li and Du] and [Martzloff].

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The Jiuzhang Suanshu (Nine Chapters on the Mathematical Art) The Jiuzhang Suanshu (Nine Chapters on the Mathematical Art) is a collection of 246 mathematical problems in nine chapters, consisting of work representative of the development of ancient Chinese mathematics from the Zhou, Qin and Han dynasties. It is the longest surviving and most influential of Chinese mathematics texts. It has many commentaries and has been translated into English, German and Russian. The 246 problems of the Jiuzhang Suanshu and their solutions show that the Chinese had access to a variety of formulas for determining areas and volumes of basic shapes, as well as methods for finding square and cube roots and solving linear equations and systems of equations in two and three unknowns. They used a decimal system of numeration, including decimal fractions. They were the first known society to use negative numbers in calculations. The Jiuzhang Suanshu is not a theoretical text but a practical handbook for architects, engineers, surveyors and tradesmen. Candidates for civil service positions studied this text, among others, as a preparation for questions on the imperial examinations. An individual's results on the civil service examinations would determine whether or not a candidate would earn a bureaucratic position in the imperial government. The 246 problems in the Jiuzhang Suanshu are arranged within nine chapters, in a question-answer format. Each section in chapter contains a general rule (shu) that enables one to arrive at a solution (chao) for the problems in that section. The problems are closely connected with practical life, and they reflect the wisdom and abilities of the ancient Chinese. Chapter 1 is entitled Fangtian (Field measurement). It discusses how to calculate areas of cultivated land, and how to do computations with fractions. Chapter 2 of the Jiuzhang Suanshu is entitled Su mi (Cereals). It focuses on proportion problems, particularly those related to the exchange of cereals. Chapter 3 is entitled Cui fen (Distribution by proportion). It discusses problems related to proportional distribution. Chapter 4 is entitled Shao guang (What width?). It discusses how to calculate side lengths when given area or volume, and how to find square roots and cube roots. Chapter 5 is entitled Shang gong (Construction consultations). It discusses various kinds of calculations for constructions, in particular, calculations involving volumes of various shapes of solids. Chapter 6 is entitled Jun shu (Fair taxes). It focuses on calculating how to distribute grain and corvĂŠe labor in the best way based on population and the distance between places. Chapter 7 is entitled Ying bu zu (Excess and deficiency). It focuses on using the method of false position to solve difficult problems. Chapter 8 is entitled Fang cheng (Rectangular Arrays). It discusses simultaneous linear equations, concepts involving positive and negative numbers, and methods of addition and subtraction of positive and negative numbers. Lastly, chapter 9 is entitled Gougu. It discusses the Gougu theorem (Pythagorean theorem), and has problems dealing with right-angled triangles. See [Crossley].

Indian (Hindu) Mathematics (1000 BCE - 1200 CE) Our earliest evidence of Indian (Hindu) mathematics dates back to the Aryan tribes who migrated from the Asian steppes and settled along the Ganges River in the latter part of the

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second millennium BCE. Around the eighth century BCE these tribes had monarchical states whose governing activities involved large scale irrigation, a centralized administration and fortifications. The social system of these states was highly stratified. The highest strata consisted of the Brahmins (Hindu priests) and the ruling class. The literature of the Brahmins was expressed in long verses called Vedas, which were transmitted from generation to generation by oral tradition. The mathematical sections of these Vedic works are known as the Sulvasutras. They are a major source of information on Indian (Hindu) mathematics. They focus on the mathematics of brick altars. However, the Vedic civilization did not have a tradition of brick technology. Interestingly, the civilization of the Harappan in the third millennium prior to the Vedic civilization which arose on the banks of the Indus River, had a tradition of brick technology. This seems to suggest that the mathematics in the Sulvasutras was originally transmitted from the Harappan civilization to the Vedic civilization. See [Joseph]. Other major sources of information on Indian (Hindu) mathematics come from the writings of Aryabhata in the sixth century, Brahmagupta in the seventh century, Mahavira in the ninth century, and Bhaskara in the twelfth century.

Greek Mathematics (800 BCE â&#x20AC;&#x201C; 800 CE) The Greek civilization emerged along the shores of the Mediterranean Sea and the Black Sea, in the wake of the decline of the ancient civilizations of Egypt and Babylon. Historical records indicate that the Greek mathematical and literary tradition is borrowed and refined from other cultures, particularly those of the Egyptians and Babylonians. The Greeks did not have a centralized governing structure. Governing was done on a polis (city-state) basis. The Greek polis were diverse in their governing nature. Almost every polis had access to the sea, resulting in continuous trade and migration between Greece and other civilizations. Greeks learned geometry from the Egyptians, and learned astronomy from the Babylonians. Greek mathematics was highly theoretical. This was probably due to the debating tradition of the Greeks. This tradition of Greek society was based on skepticism and demonstrative evidence. It led the Greeks to develop the nature of proof in mathematics and to consider differing solutions to fundamental questions about the world. The major texts of Greek mathematics, those by Euclid, Archimedes, and Apollonius, written around the third century BCE, deal primarily with geometric concepts. Although one can find indications that Greek mathematicians of that time had some ideas about solving quadratic and cubic equations, they did not develop arithmetic algorithms to do so. Later on, however, two mathematicians in the Greek tradition living in Alexandria, Heron (1st century CE) and Diophantus (3rd century CE), did develop some arithmetic and algebraic techniques to solve equations. See [Heath] and [Cuomo].

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Islamic (Arabic) Mathematics (700 CE - 1200 CE) The Islamic world was fertile ground for the development of mathematics and the preservation of ancient mathematical traditions. Baghdad (the capital of Iraq) established in 766 CE was the center of learning for the Islamic world. By the ninth century CE, its recognition as an intellectual center was achieved with the establishment of a library and research center called Bayt al-Hikma (House of Wisdom), by the Caliph al-Ma'mun (reigned 813 CE – 833 CE). This center attracted scholars from all over the Islamic world. These scholars conducted their own research and assisted with translating documents collected by the library, from all parts of the Islamic world, into Arabic. Manuscripts collected included many classical Greek texts by scholars from Athens and Alexandria (e.g., Euclid and Diophantus). Among these Islamic scholars was Abu-Abdullah Mohammed ibn-Musa Al-Khwarizmi, a great mathematician and astronomer. See [Berggren]. Abu-Abdullah Mohammed ibn-Musa Al-Khwarizmi (c 780 CE – 847 CE) Abu-Abdullah Mohammed ibn-Musa Al-Khwarizmi (Mohammed, father of Abdullah and son of Moses, the Khwarezmite) is considered to be one of the greatest Islamic mathematicians of his time. He came from Khwarizm, an area which is now part of Uzbekistan. He was one of the many scholars that the Caliph al-Ma'mun gathered together in Baghdad's House of Wisdom. Around 820 CE, Al-Khwarizmi wrote his famous book, Arithmetic. It was the first Arabic text to deal with the Hindu numerals. These numerals probably arrived at the court of the Caliph in Baghdad in 773 CE with the diplomatic mission from Sind to the court of al-Mansur. Around 825 CE, Al-Khwarizmi wrote another famous book Al-kitab al-muhtasar fi hisab al-jabr w’al’muqabalah (The Condensed Book on the Calculations of al-jabr and al-muqabala), from which the term algebra is derived. This book impacted the process of mathematical calculations in places as distant as Europe for many centuries. It is commonly referred to, in its shortened form, as Al-jabr. The second half of this text is devoted to problems dealing with Islamic inheritance law. Among the laws are that when a woman dies her husband receives 1/4 of her estate, and the rest is divided among her children such that a son receives twice as much as a daughter, and that strangers cannot receive more than 1/3 of the estate without permission of the heirs. These laws form the basis for some interesting mathematical problems. In addition to his famous mathematics books, Al-Khwarizmi also wrote on the sundial and the works of Ptolemy. He was familiar with the mathematics and astronomy of the Hindus. `Umar ibn Ibrahim al-Khayyami (1048–1131) Al-Khayyami (often known as Omar Khayyam), was born in Nishapur, now in Iran, shortly after the area was conquered by the Seljuk Turks. He was able during most of his life to enjoy the support of the Seljuk rulers. In fact, he spent many years at the observatory in Isfahan at the head of a group working to reform the calendar. At various times, as ruler replaced ruler, he fell into disfavor, but he was able ultimately to garner enough support to write many

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mathematical and astronomical works, as well as poetry and philosophical works. In fact, he is best known in the West for the collection of poems known as the Rubaiyat. A story is told that as a student, Omar Khayyam made a compact with two fellow students, Nizam al Mulk and Hassan ibn Sabbah, to the effect that the one who first achieved a high position and great fortune would help the other two. It was Nizam who in fact became the grand vizier of the Seljuk Sultan Jalal al-Din Malik-shah and proceeded to fulfill his promise. Hassan received the position of court chamberlain, but after he attempted to supplant his friend in the sultanâ&#x20AC;&#x2122;s favor, he was banished from the court. Omar, on the other hand, declined a high position, accepting instead a modest salary which permitted him to have the leisure to study and write.

Western Mathematics (1200 CE â&#x20AC;&#x201C; Present) Western mathematics took center stage towards the end of the Medieval period. The late twelfth and thirteenth centuries marked a period of growth with the establishment of famous universities, for example, Bologna, Paris, Oxford, Cambridge, and Gothic cathedrals, for example, Chartres, Notre Dame, Westminster, Rheims. The mathematical works of prior civilizations, including the Greek and Islamic, were translated into Latin, making them easily accessible to Western mathematicians. Europeans learned about solving quadratic equations from the work of al-Khwarizmi, both in Latin translation from the Arabic during the twelfth century as well as through the Hebrew work, Treatise on Mensuration of Abraham bar Hiyya (d. 1136), later translated into Latin, and the very detailed Liber Abbaci by Leonardo of Pisa (editions in 1202 and 1228). (Leonardo is more familiarly known as Fibonacci.) Although Islamic mathematicians attempted to solve cubic equations, and were successful in solving them geometrically, it was not until the sixteenth century in Italy that an algebraic solution was found. Algebraic solutions to fourth degree polynomial equations were also discovered at that time. For centuries after that, mathematicians sought to find similar algebraic solutions for polynomial equations of degree five and higher. However, although it was certainly possible to solve some equations of that type, Niels Henrik Abel in the mid 1820s finally proved that it was impossible to find an algebraic method of solving fifth degree polynomial equations in general (or, of course, polynomials of any higher degree). Shortly thereafter, Evariste Galois worked out group-theoretic methods of determining whether, in fact, a given polynomial equation could be solved algebraically. Galois theory, as his ideas are now called, led to many new advances in algebra in the nineteenth and twentieth century. Brief biographical sketches of some of the major figures in the development of methods for solving polynomial equations follow. More extensive biographies can be found at the St. Andrews website: http://www-history.mcs.st-and.ac.uk/history/

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Nicolas Chuquet (1445–1488) Nicolas Chuquet was a French physician, who wrote his mathematical treatise, the Triparty, in Lyon near the end of his life. He wrote the book, a work on arithmetic and algebra in three parts, to meet the need for practical mathematics of the thriving commercial community of Lyon In the first part of his book, he introduced his fraction rule which is discussed in this module. In the second part, he applied his rule to find the roots of polynomials and to calculate the square and cube roots. In the third part, he introduced several types of equations and methods of solutions. He then discussed numerous problems in various fields to which his rules could be applied. Niccolo Tartaglia (1499–1557) Tartaglia was one of the first mathematicians to discover algebraic solutions to cubic equations. In fact, he displayed this knowledge in a public contest with Antonio Maria Fiore, who had learned of one particular solution from his mentor, Scipione del Ferro. Tartaglia was eventually persuaded by Cardano to reveal some of his methods, on the promise that Cardano would not publish them. However, for various reasons Tartaglia never published his solutions and Cardano eventually decided that he would do so. Tartaglia became very bitter and, although he published a useful work on the mathematics of artillery, he died a broken man. Gerolamo Cardano (1501–1576) Cardano was trained as a physician and eventually became the most prominent physician in Milan. His most important patient was probably John Hamilton, the Archbishop of Scotland, who in 1551 requested Cardano’s services to help him overcome steadily worsening attacks of asthma. Cardano, after observing the archbishop’s habits for a month, decided he had a severe allergy to the feathers in his bed. Thus, Cardano recommended that the bedding be changed, and the archbishop’s health immediately improved. Cardano wrote numerous texts during his life, including several important ones on mathematics. His Ars Magna (The Great Art, or On the Rules of Algebra) was primarily devoted to methods of solving cubic equations. It was extremely influential in creating a new attitude toward algebra in Europe. An image of the cover page of the Ars Magna is included below. It could be posted on a bulletin board or made into an overhead transparency. Robert Recorde (1510–1558) Recorde graduated from Oxford University in 1531 and then licensed in medicine. Although he probably practiced medicine in the 1540s, he is best known for writing several successful mathematics textbooks, including The Ground of Arts on arithmetic, The Whetstone of Witte on algebra, and The Pathway to Knowledge on geometry. His works show that he was especially interested in pedagogy. In particular, his books were set in the form of a dialogue between master and pupil, in which each step in a particular technique was carefully explained. An image of the cover page of The Whetstone of Witte is included below.

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Lodovico Ferrari (1522–1565) Ferrari was a pupil of Cardano’s who lived and worked with him for many years. He assisted Cardano in the studies leading up to Cardano’s full explication of solutions of cubic equations in the Ars Magna. During that period, he discovered how to solve fourth degree polynomial equations algebraically, and Cardano discussed his methods in his text. Ferrari also was Cardano’s spokesperson in his conflict with Tartaglia, defeating the latter in a public mathematics contest in 1547. Rafael Bombelli (1526–1572) Bombelli was educated as an engineer and spent much of his adult life working on engineering projects in the service of his patron, a Roman nobleman. He worked on mathematics during lulls in his engineering work. His aim in writing an algebra text, which he completed by 1560, was to clarify and expand Cardano’s work in the Ars Magna. In particular, he showed why complex numbers were important in using Cardano’s formula to solve certain cubic equations. An image of the cover page of Bombelli’s Algebra is included below. François Viéte (1540–1603) Viète was born in Fontenay-le-Comte, Poitou (now Vendée), France. He was trained as a lawyer and served King Henri III and his successor Henri IV as an advisor. One of his tasks, in fact, was to act as a cryptanalyst of intercepted messages between the king’s enemies. Although mathematics was only an avocation, he contributed greatly to the development of algebra, being the first to use letters to represent constants in algebraic expressions. His collected works, entitled The Analytic Art, have earned him the title of “Father of Algebra.” He died in Paris in 1603. René Descartes (1596–1650) Descartes was born at La Haye (now La Haye-Descartes) near Tours, France into a family of the old French nobility. Because he was sickly throughout his youth, he was permitted during his school years to rise late. He thus developed the habit of spending his mornings in meditation. Because his thoughts led him to the conclusion that little he had learned in school was certain, he became so full of doubts that he abandoned his studies to spend time traveling and learning from all sorts of people. In 1628, he settled in the Netherlands to begin his lifelong goal of creating a new philosophy suited to discovering truth about the world. He explained some of his ideas in the 1637 work, Discourse on the Method for Rightly Directing One’s Reason and Searching for Truth in the Sciences. This work was accompanied by three essays giving examples of Descartes’s ideas: Geometry, Optics, and Meteorology. In 1649, he was invited by Queen Christina of Sweden to come to Stockholm to tutor her. However, his health could not withstand the severity of the northern climate, especially because Christina required him, contrary to his long-established habits, to rise at an early hour. Descartes soon contracted a lung disease which led to his death in 1650.

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Pierre de Fermat (1601–1665) Fermat was born into a moderately wealthy family in Beaumont-de-Lomagne in the south of France. He studied at the University of Toulouse and took a law degree at Orleans in 1631. He spent most of the rest of his life in Toulouse, where he was a jurist, serving for many years in the chambers of the Parlement, a body charged with both administrative and legal functions. Fermat considered mathematics as a hobby, a refuge from the continual disputes with which he had to deal as a lawyer. He therefore refused to publish any of his discoveries, because to do so would have forced him to complete every detail and to subject himself to possible controversies in another arena. Nevertheless, many of his manuscripts with sketches of his ideas circulated in the European mathematical community and proved immensely influential. Isaac Newton (1642–1727) Newton was born on Christmas Day, 1642 at Woolsthorpe, near Grantham, some 100 miles north of London, to a mother already widowed in October. After his mother remarried, she left young Isaac in the care of his grandmother, before sending him away to grammar school in Grantham in 1655. After mastering the curriculum at Grantham, he matriculated at Trinity College, Cambridge in 1661. During his years at Cambridge, Newton explored on his own all the new discoveries in mathematics in the first half of the seventeenth century, and then, using his immense powers of concentration, proceeded to develop these ideas further into the subject of calculus. After being appointed Lucasian Professor of Mathematics in 1669, he carried on studies in fields as diverse as physics and religion, eventually publishing his master work, the Mathematical Principles of Natural Philosophy, in 1687. During the last three decades of his life, Newton served as Master of the Mint in London as well as President of the Royal Society. Joseph-Louis Lagrange (1736–1813) Born in Turin to a family of French descent, Lagrange was attracted to mathematics from an early age and became a professor at the Royal Artillery School in Milan when he was only 19. He soon impressed Leonhard Euler with his mathematical prowess and, when Euler left the Berlin Academy of Sciences to return to St. Petersburg, Frederick II of Prussia appointed Lagrange to the vacant post. After Frederick’s death, Lagrange accepted the invitation of King Louis XVI of France to come to Paris, where he remained for the rest of his life. His most important work, Analytical Mechanics, was published in France in 1788. After the French Revolution, Lagrange took an active role in improving university education in France. Niels Henrik Abel (1802–1829) Born near Stavanger in Norway, Abel’s native abilities in mathematics were discovered by his instructor at the Cathedral School in Oslo, who encouraged Abel to read various advanced mathematics treatises available at the university. During his later university studies, he discovered the impossibility of solving fifth degree equations algebraically. He also made major strides in the theory of elliptic functions. Although he received a grant to travel and study in Europe after he graduated, he found on his return that there were no available university positions in his country. He struggled to make a living by tutoring and substituting at the

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university, meanwhile preparing a large number of new mathematical articles. But in January, 1829, he suffered an attack of tuberculosis from which he was unable to recover. He died in April, two days before August Crelle wrote to him with the news that an appointment had been secured for him in Berlin. Evariste Galois (1811–1832) Galois’s tragically brief life has been the subject of a fictionalized biography which included speculation that his death in a duel was engineered by government agents because of his radical political views. The known facts, however, do not support this contention. Nevertheless, although Galois introduced many new ideas into algebra beginning at about age 18, he did not achieve the recognition he desired, partly because the mathematicians at the French Academy could not understand his concepts. After the July revolution of 1830, Galois was arrested for threatening the life of the king. Although he was soon released from prison, he somehow became involved with “an infamous coquette and her two dupes,” and was forced to fight a duel in which he was killed, five months before his twenty-first birthday. On the night before the duel, expecting to die, he wrote a letter to his friend Auguste Chevalier amplifying and annotating some of his earlier manuscripts. It was not until about fifteen years later that his work was eventually understood and published. Translation of the Title Page of the Ars Magna THE GREAT ART or The Rules of Algebra by GIROLAMO CARDANO, Outstanding Mathematician, Philosopher and Physician In One Book, Being the Tenth in Order of the Whole Work on Arithmetic Which is called the Perfect Work. In this book, learned reader, you have the rules of algebra (in Italian, the rules of the coss). It is so replete with new discoveries and demonstrations by the author – more than seventy of them – that its forerunners are of little account or, in the vernacular, are washed out. It unties the knot not only where one term is equal to another or two to one but also where two are equal to two or three to one. It is a pleasure, therefore, to publish this book separately so that, this most abstruse and unsurpassed treasury of the entire subject of arithmetic being brought to light and, as in a theater, exposed to the sight of all, its readers may be encouraged and will all the more readily embrace and with the less aversion study thoroughly the remaining books of the Perfect Work which will be published volume by volume.

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Finding Square Roots Teacher Notes Level: Parts of this activity can be used in middle school; other parts in high school algebra courses. Materials: Historical background; student pages; calculator Time Frame: Each part of this activity can be done in one class period. You should only use a few of these parts with a given class. Objectives: To demonstrate hand methods of calculation of square roots. To give students an introduction to the notion of a convergent sequence. When to Use: The only prerequisite for most of this material is basic arithmetic. However, many of the calculations necessary can be done on a calculator. How to Use: Explain to the class that you will be considering several methods for solvN , for N a positive number. Although one could ing the simple quadratic equation x2 = √ √ express the solution to this equation as N , or as ± N , what we want to do here is find a numerical approximation to that solution. Each part of this activity develops a different method for finding the numerical approximation. They can be worked on by students in small groups. It may be useful to assign a different method to different groups and have the groups compare results afterwards.

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Finding Square Roots Student Pages Historical Background Finding square roots has been of interest to people as far back as documentation exists. In ancient times, once people understood that the area of a square could be calculated by multiplying the length of a side by itself, they tried to solve the inverse problem. Namely, if they knew the area of a square, they wanted to determine the length of a side. This is precisely the problem of finding a square root. A more natural word for square root, one that was in fact used by the ancient Babylonians, is “square side”, that is, the side of a particular square. In ancient Egypt, as far as we can determine, the scribes wrote down tables of squares of various numbers. When they needed to find a square root, they simply used the table in reverse. They looked down the list of squares to find the desired number and then found the side which gave that square. Of course, we know that this method does not always work. No matter how detailed your table, there will be missing numbers. What we do not know, as far as ancient Egypt is concerned, is how the scribes proceeded in that case. In the documents that are available (and there are very few from ancient Egypt) the scribe simply wrote down the square root for a given square without giving any indication of how the value was found. In ancient Babylonia, tables were used as well. However, there is indication on some of the tablets of a method whereby the scribes were able to determine square roots. This method is based on geometry, namely on dividing up the square in certain ways. We will consider that method in one of the activities below. From China, during the Han period (approximately 200 BCE – 200 CE), there is explicit documentation of an algorithm for determining square roots. Scholars believe that this algorithm also comes from geometry. We will look at this method as well. Archimedes, around 250 BCE, used a purely numerical algorithm to calculate square roots, one very similar to the Babylonian one. Ptolemy, around 150 CE, also used an algorithm, one slightly different, which was explicitly documented by Theon two centuries later. Heron, another Greek mathematician living in Alexandria in the first century CE, gave an entirely different algorithm. Other numerical algorithms were developed in medieval China and in medieval Islam. Further algorithms were developed in the Renaissance by, among others, Nicolas Chuquet and Fran¸cois Vi`ete. We will look at some of these algorithms as well.

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Babylonian Square Root Method Although most of the time, square roots on Babylonian tablets are simply written down, there is some evidence that the scribes used a geometric method based on the identity (x + y)2 = x2 + 2xy + y 2 . Although the Babylonians calculated in base 60, we will modify their method so √ we just use ordinary fractions. Now given a square of area N for which one wants the side N , the first step would be to choose a number a (whole number or fraction) close to, but less than, the desired result. Setting b = N − a2 , the next step is to find c so that 2ac + c2 is as close as possible to b (See figure). If a2 is “close enough” to N , 2 then c2 will be small in relation to 2ac. Therefore, √ ignore the c term and simply √ we can solve 2ac = b. Thus c is equal to b/2a. That is, N = a2 + b ≈ a + b/2a. (Note that if a is chosen √ initially to be larger than the desired result, a similar geometric argument shows that a2 − b ≈ a − b/2a.

√ We now have two approximations to N . The first one is r1 = a; the second is r2 = a ± b/2a, with the sign choice depending on whether the first approximation is two small or two large. If we want a more accurate approximation, we can repeat this procedure. That is, set b0 = N − r22 or b0 = r22 − N and find a new value for c by the formula: c0 = b0 /2r2 . Then our next approximation is r3 = r2 ± b0 /2r2 . We can continue in this way for as many steps as we want. What this means is that we are finding a sequence of numbers, r1 , r2 , r3 , . . . which gets closer and closer to a given value, in this case the actual length of the side of the square of area N . This idea of choosing a sequence which approaches a given value will be extremely important in your later studies of mathematics. Examples: √ 1. Let us approximate 10 by this method. Our first step is to pick a number a whose square is close is a = 3. So r1 = 3 is our first approxi√ to 10. One convenient value 2 mation to 10. Next, we set b = 10 − 3 = 1 and set c = b/2a = 1/6. Then our second approximation is r2 = 3 + 1/6 = 3 1/6. We check that (3 1/6)2 = 10 1/36, so r2 is a bit too large. If we want a better approximation, we repeat the process. Set c 2004, M.A.A.

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b0 = (3 1/6)2 − 10 = 10 1/36 − 10 = 1/36. Then set c0 =

1/36 1 b0 1/36 = = . = 2r2 2(3 1/6) 6 1/3 228

Therefore our next approximation is r3 = 3 1/6 − 1/228 = 721/228 = 3 37/228. The square of 3 37/228 is 1 519, 841 = 10 . 51, 984 51, 984 Thus the square of this approximation only differs √ from 10 by 1/51, 984, or 0.0000192. To put this another way, the calculator value of 10 is 3.16227766, while our value r3 is equal, in decimals, to 3.162280702. Thus, these values differ by approximately 0.000003, or by approximately 0.0001%. This is certainly accurate enough for most purposes, but if you want even better accuracy, you can do another step in the procedure. √ 2. We approximate 3. The first step is to pick a starting number a = r1 whose square is close to 3. We could pick r1 = 2, but instead we will start with a fraction r1 = 1 3/4. Then, since the square of r1 is larger than 3, we set b = (1 3/4)2 − 3 = 1/16. Thus, c = b/2a = (1/16)/(7/2) = 1/56, and our second approximation is r2 = 1 3/4 − 1/56 = 97/56 = 1 41/56. Using your calculator, you will see that this value agrees with the square root of 3 to 3 decimal places. But we will improve this by taking one more step in the algorithm. We calculate that (1 41/56)2 > 3, so we set b0 = (1 41/56)2 − 3 = 1/3136. Then 1 b0 1/3136 c0 = = , = 2r2 2(1 41/56) 10864 and our next approximation is r3 = 1 41/56 √ − 1/10864 = 1 7953/10864 ≈ 1.73205081, which agrees with the calculator value of 3 to 8 decimal places. Exercises: √ 1. Approximate 2 to at least five decimal places, using the Babylonian √ method. Start with r1 = a = 1 1/3. How close is r3 to your calculator’s value for 2. √ 2. Approximate 3 using the Babylonian method, but this time start with r1 = a = 2. Compare your result for r3 here to the corresponding result in example 2. √ 3. Approximate 65 to five decimal place accuracy, using the Babylonian method. Start with a = r1 = 8. √ 4. Approximate 7 to five decimal place accuracy, using the Babylonian method. Do this at least two ways, first starting with a = r1 = 5/2 and then starting with a = r1 = 8/3. Compare your results.

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Chinese Square Root Algorithm The Chinese method, described in detail in the Nine Chapters of the Mathematical Art, a classic work written near the beginning of our era during the Han dynasty, is also based on the algebraic/geometric rule (x+y)2 = x + 2 xy + y2. This method was explicitly designed for a decimal system, and is essentially the method that was typically taught in American schools in the era before calculators. It is designed to give a series of approximations to the square root of a number, each approximation having one more significant figure than the previous one. Because basic arithmetic procedures in the decimal system do not depend on the position of the decimal point, this method allows one to calculate as many decimal places as desired for the square root, one decimal place at a time. We illustrate this algorithm by problem 12 of chapter 4 of the Nine Chapters, where we are asked to determine the side of a square of area 55,225. Chinese mathematics used the decimal system. Thus, the idea is to find digits a, b, c so that the answer can be written as 100a + 10b + c.

First find the largest digit a so that (100a)2 < 55225. In this case, a = 2. The difference between the large square (55,225) and the square on 100a (40,000) is the large L-shaped figure in the diagram. (Such an L-shaped figure is usually called a gnomon.) If the outer thin gnomon is neglected, we see that b must satisfy 55,225 – 40,000 > 2(100a)(10b) or 15,225 > 4000b. So certainly b < 4. To check that b = 3 is correct, that is, that with the square on 10b included, the area of the large gnomon is still less than 15,225, it is necessary to check that 2(100a)(10b) + (10b)2 = 10b [2(100a) + 10b] = 10b(400 + 10b) < 15,225. Because this is in fact true, the same procedure can be repeated to find c: 55,225 – 40,000 – 30(2 × 200 + 30) > 2 × 230c or 2325 > 460c. Evidently, c < 6. An easy check shows that c = 5 gives the correct square root: 55, 225 = 235 . In this particular case, the square root is an integer. If it had not been an integer, we could have continued the process using decimals. Although we have explained the Chinese procedure geometrically, in fact the Chinese presented it as an arithmetic algorithm. The algorithm starts by determining the number of integral places the square root has. To do this, it is simplest to break up the number into pairs, beginning at the decimal point and moving left. In the case of 55,225, we break it up as 5 52 25, with only a single digit in the leftmost place. That there are three groups of digits shows that

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the square root will have three integral places. (Why?) Next, one finds the largest integer whose square is less than or equal to the left most group of digits. In this case, 2 is that integer, because 22 = 4 and no larger square is less than or equal to 5. Thus, we know that the square root is a three digit number beginning with 2. In other words, 200 is our initial approximation to the square root of 55,225. For our next step, we subtract 2002 = 40,000 from 55,225 to get 15,225. In performing this algorithm, however, we often just think of this step as subtracting 22 = 4 from the left most group of digits, namely 5. This leaves 1 for that digit â&#x20AC;&#x201C; but of course this 1 stands for 10,000 and we still have an additional 5,225 in the given number. _2__________ | 5 52 25 _4______ 1 52 In the next step, we double the 2, giving 4, because we have two rectangles each of which has that as one side, then guess the digit b that we can place next to the 4 and also next to the 2 in the answer line at the top, so that when we multiply the 4b (which stands for 400 + 10b) by the b, the result is no greater than the 152 already there. In this case, the correct value is 3. We then multiply, giving 129, subtract that value from 152, leaving 23, and bring down the final pair to give 2325. _2__3________ | 5 52 25 _4______ 43 | 1 52 1_29___ 23 25 We have now determined that the first two digits of our square root are 23; that is, our second approximation to the square root of 55,225 is 230. We now repeat the procedure. We double the 23, giving 46, and figure out what digit c we can place next to both the 46 and the 23 in the answer line so that when we multiply c by 46c (=2(200 + 30) + c) the result is no greater than 2325. In this case, the correct value if 5, and when we multiply 465 by 5, we get 2325. Thus, our square root is an integer, 235. _2__3__5______ | 5 52 25 _4______ 43 | 1 52 1_29___ 465 | 23 25 _23_25 0

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We needed three approximations to arrive at our final result: 200, 230, 235. In this case, we achieved our result exactly. Now, let us look at an example where we cannot get an exact answer. We calculate 3 , where we decide in advance how many decimal places we desire. Suppose we want three places after the decimal point. So we rewrite 3 as 3.00 00 00, where each pair of 0s will give one digit in the answer. This answer will have a single digit to the left of the decimal point and three digits to the right. But as usual, once we have determined where the decimal point will lie in the answer, we proceed just as in the integral case. We begin, as before, by determining the largest integer whose square is less than or equal to 3, the leftmost digit in our problem. That number is 1. So we square 1 to get 1 and subtract that value from the 3, leaving 2 followed by our three pairs of 0s. _1__________ | 3 00 00 00 _1______ 2 00 00 00 Next, we double the 1 to get 2 and guess a digit b, which we place next to the 1 in the answer line, such that b multiplied by the number 2b is as large as possible but still less than 2 00. After some experimenting, we find that the correct value for b is 7. We multiply 7 by 27 to get 189, then subtract from 200 and bring down the 0s: _1__7________ | 3 00 00 00 _1______ 27 | 2 00 00 00 1_89___ 11 00 00 We continue in the same manner. We double the 17 to get 34 and guess a digit c to place next to the 17 in the answer line and next to the 34 so that c multiplied by 34c is as large as possible but still less than 11 00. The correct value is 3. We multiply 3 by 343 to get 1029, then subtract from 1100 and bring down the 0s again. _1__7__3______ | 3 00 00 00 _1______ 27 | 2 00 1 89___ 343 | 11 00 00 10_29_ 71 00

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To get one more place in our answer, we double the 173 to get 346 and find the digit d so that d multiplied by 346d is as close to 7100 as possible. In this case, the correct value is 2. Again, we multiply and subtract. In contrast to our earlier example, we do not get an exact answer. In this case, we will simply stop at this point, with an answer of four significant digits. Given that the final remainder is small relative to the final divisor, we will leave the final digit as it is. We also know that the decimal point must be placed after the 1, so that our approximation for 3 is 1.732. _1__7__3__2____ | 3 00 00 00 _1______ 27 | 2 00 1 89___ 343 | 11 00 00 10_29_ 3462 | 71 00 69_24 1 76 As before, we have calculated a sequence of numbers, 1, 1.7, 1.73, 1.732, each of which is a better approximation to 3 than the one before. Since 3 is an irrational number, there can be no exact decimal representation. The best we can hope for is such a sequence of better and better approximations. With this Chinese method, each step just gives us one more decimal place. Thus, if we decide in advance how many decimal places we want, we will know how many steps to take using this method. Exercises: Use the Chinese method to calculate the square roots of each of the following. In each case, carry out the calculation to four significant figures (unless there is an exact answer in fewer figures). 1. 127,449 2. 4,004,001 3. 17 4. 233.5 5. 1,234.5678

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Heron’s Method Heron was a mathematician who lived in Alexandria, Egypt during the first century C.E. He wrote several books dealing with the applications of mathematics to such fields as optics and mechanics. Although he did give proofs of some of his results, he was more interested in getting answers, so sometimes he just described methods which he knew would work. For example, in his work entitled Metrica, he instructed his readers how to determine areas and volumes of various types of figures. He was generally interested in actual numerical results. Thus, if his procedure involved a square root, he had to show how to determine one. We present here a quotation from Heron describing his method of finding the square root of 720. Since 720 has not a rational square root, we shall make a close approximation to the root in this manner. Since the square nearest to 720 is 729, having a root 27, divide 27 into 720; the result is 26 23 ; add 27; the result is 53 23 . Take half of this; the result is 26 56 . Therefore the square root of 720 will be very nearly 26 56 . 1 1 For 26 56 multiplied by itself gives 720 36 ; so that the difference is 36 . If we wish 1 to make the difference less than 36 , instead of 729 we shall take the number now 1 found, 720 36 , and by the same method we shall find an approximation differing by 1 much less than 36 . Heron’s procedure is often called the ”divide and average” procedure. Namely, we begin with an approximation to the square root of a number. We divide this approximation into the number. We next average our first approximation with this quotient to give a second approximation. We then repeat the procedure by dividing our second approximation into the number. We average this quotient with the second approximation to give a third approximation. As in our other procedures, we can carry out this process as many times as we want. At each stage we get a closer approximation than at the previous stage. √ To give another example, let us approximate 2 using Heron’s procedure. We begin with a1 = 43 as our first approximation. The quotient q1 of 2 by 43 is 32 . So the second approximation a2 is the average of 43 and 32 . Thus,

1 4 3 17 + = . a2 = 2 3 2 12 Note that the square of 17/12 differs from 2 by only 1/288. Since the difference between a21 = (4/3)2 and 2 is 2/9, we see how much better our second approximation is than our first. But if we are not satisfied with this value, we can run through the procedure again. We divide 2 by 17/12, giving 24/17. Thus, our third approximation is a3 =

1 2

17 24 + 12 17

577 . 408

If we square 577/408, we get 332, 929/166, 464, which differs from 2 by only 1/166, 464, or approximately 0.00001. We will stop here, although we could continue the approximation for several more steps if desired. c 2004, M.A.A.

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Exercises: √ 1. Carry out the next repetition of Heron’s method in the calculation of 720. What is the percentage error in your final value, a3 , compared with the true value as found on your calculator. √ 2. Carry out one further step in Heron’s method in the calculation of 2. Again, determine the percentage error in using a4 . Use Heron’s method to find the square roots of each of the following numbers. Carry out your calculations far enough so that your answer differs from the true one by less than 0.001%. 3. 3 4. 8 5. 1234.5

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Chuquet’s Method Nicolas Chuquet (d. 1487) was a French physician who worked in Lyon. Near the end of his life, he wrote a mathematical treatise, the Triparty, designed to meet the needs of the thriving commercial community in Lyon. Thus, he included many practical problems in his text on arithmetic and algebra. One of the problems with which Chuquet dealt is the finding of square roots of nonsquare numbers. He first noted that to find a fraction between two fractions, one just needs to add the numerators and the denominators of the given fractions. Thus, between 1/2 and 1/3 is 2/5, and between 1/2 and 2/5 is 3/7. Exercise: Show why, in general,

a a+c c < < , b b+d d

where a, b, c, d are positive integers. Chuquet then applied this rule to the calculation of square roots. For example, to calculate the square root of 6, he noted that 2 is too small and 3 is too large. Then he checked that 2 1/3 is also too small and 2 1/2 is too large. Thus the desired root is between these two numbers. So he used his procedure to find a number “between.” That is, he picked 2 2/5. He then checked whether the square of 2 2/5 is smaller or larger than 6. Since (2 2/5)2 < 6, his next choice was a number between 2 2/5 and 2 1/2, namely 2 3/7. Again checking the square, he found that (2 3/7)2 < 6, so he needed a new number between 2 3/7 and 2 1/2. According to his procedure, this next value is 2 4/9. Exercises: 1. Continue in this manner. First, check whether the square of 2 4/9 is smaller or larger than 6. If it is smaller, use Chuquet’s method to find a number between 2 4/9 and 2 1/2. If it is larger, then use the method to find a number between 2 3/7 and 2 4/9. Continue in this manner for several more steps. Show that you eventually reach the number 2 89/198. Show that the square of 2 89/198 differs from 6 by 1/39, 204. (Chuquet decided that this approximation was close enough.) 2. Use Chuquet’s method to find a series of approximations to the square root of 2. Start with the two values 1 1/4, whose square is less than 2, and 1 1/2, whose square is greater than 2. Stop when your answer is correct to five decimal places. 3. Use Chuquet’s method to find an approximation to the square root of 5 accurate to five decimal places.

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Newton’s Method One of Isaac Newton’s earliest mathematical ideas was a method for solving equations numerically, a method he adapted from the work of Fran¸cois Vi`ete. We explore this method in√ the case where the equation is the pure quadratic equation x2 − 5 = 0, that is, in finding 5. Newton’s idea was to begin with a reasonable guess of the solution for our initial approximation. Here we will take x1 = 2. Then, we set x = 2 + p, where p is small and try to figure out a value for p to give our second approximation. To do this, we substitute 2 + p for x in our equation: (2 + p)2 − 5 = 0 and expand. We get 4 + 4p + p2 − 5 = 0, or p2 + 4p − 1 = 0. Rather than trying to solve this quadratic equation exactly, Newton used the fact that because p is small, p2 will be much smaller and could be neglected in getting a new approximation. Therefore, he replaced that quadratic equation with the linear equation 4p − 1 = 0, which he could solve and get p = 14 . Therefore, the second approximation is x2 = 2 + 14 = 2.25. To get better approximations, we repeat the process. Namely, we set p = substitute this expression into our equation for p. We get

1 +q 4

1 + q − 1 = 0 or +4 4

1 4

+ q and

1 1 1 9 + q + q 2 + 1 + 4q − 1 = 0 or q 2 + q + = 0. 16 2 2 16

Again, assume that q is “small”, so we can neglect powers of q higher than the first. We 1 = 0, an equation we solve as thus replace this equation with the linear equation 92 q + 16 √ 1 1 1 q = − 72 . Our third approximation to 5 is then x3 = 2 + 4 − 72 = 2 17 72√= 2.236111 . . .. We 2 note that (x3 ) = 5.000193, so we have quite a good approximation to 5. 1 However, if we want a still better approximation, we repeat the process. Set q = − 72 + r and substitute this in the equation for q. Once you do this, simplify, and neglect the 1 term in r2 , the resulting linear equation is 161 36 r = − 5184 . (Please check this.) Then r = √ 1 1 1 5473 and the fourth approximation to 5 is x4 = 2 + 14 − 72 − 23,184 = 2 23,184 = − 23,184 2 2.236067978 . . .. Since (x4 ) = 5.000000002, we will stop here. As we will see in another section of this module, Newton’s method can be applied to solving more general polynomial equations.

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Exercises: Use Newton’s method to calculate the square roots of the following numbers, to an accuracy of at least 5 decimal places. 1. 2 2. 17 3. 1,000 √ 4. Calculate 7 using Newton’s method in two ways. Do it first with an initial guess of x1 = 2 and then with an initial guess of x1 = 3. Does the initial guess make a difference in the number of steps needed to get 5-place accuracy? √ 5. Repeat exercise 4 for 50. Start with x1 = 7 and also with x1 = 8. √ 6. Redo the initial steps in the calculation of 5, beginning with x1 = a. Show that

1 5 a+ . x2 = 2 a In other words, show that Newton’s method is essentially the same as Heron’s method, discussed in an earlier section of this module.

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Answers to Exercises Babylonian Method

Heron’s Method

1. 1.41421

1. a3 = 26 1609 1932 = 26.83281574

2. 1.73205

2. a4 = 1 195,025 470,832 = 1.414213562

3. 8.06226

3. a3 = 1 571 780 = 1.7320513

4. 2.64575

4. a3 = 2 169 204 = 2.828431

Chinese Method 1. 357

5. a3 =

48,392,761 1,377,320

= 35.13545218

Chuquet’s Method

2. 2001

239 2. Final values are 1 169 408 and 1 577

3. 4.123

72 3. 2 305

4. 15.28 5. 35.14

Newton’s Method 1. x3 = 1 239 577 65 2. x3 = 4 528 = 4.123106

3. x3 = 31 151,437 243,164 = 31.62278

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Deriving the Quadratic Formula Teacher Notes Level: This activity is suitable in an introductory high school algebra class. Materials: Paper and pencil. Large geometric squares and rectangles as indicated. Time Frame: At least two class periods. Objective: To construct the quadratic formula, based on the geometrical principles by which it was originally discovered in ancient Mesopotamia. When to Use: This provides as introduction to the quadratic formula. It can be used to introduce the solution of quadratic equations, even before one deals with factoring. It allows you to connect the study of quadratic equations to geometry. How to Use: This activity is designed to be done in small groups. The student pages give the students the basic questions which need to be discussed and answered. Each group should have large squares and rectangles, which can be cut up if necessary. After each group has worked through to an answer in activities 1 and 3, a student from each group could be selected to present the answer to the entire class. In activity 2, two groups should exchange solutions, as noted. In activity 4, the standard form of the quadratic formula is presented. But students may need algebraic help in deriving it from the special case derived in activity 3. In fact, this may well provide an opportunity to review manipulations with radicals. It is important to emphasize that a quadratic equation is an equation about squares and rectangles – about geometric objects. It is not just an algebraic statement. The derivation here helps make that clear. Although exercises are presented at the end, there will be many more in your textbook.

Answers to Exercises: 1. 2. 3. 4.

3, 5 1, 3 3, 8 5 ± 20 5. 9 / 2 ± 53 / 2 6. 2, 4 7. No real solutions. This occurs when (b / 2)2 − c < 0 .

8. 9. 10. 11. 12. 13. 14. 15.

5, –9 −4 ± 7 3, –8 No real solution −2 ± 7 –1, –3/2 –1, –5/3 –4, –1/3

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Deriving the Quadratic Formula Student Pages Historical Background: Techniques for solving what we call quadratic equations were discovered at least 4000 years ago in ancient Mesopotamia, in the area which is now Iraq. The methods probably come from techniques of surveyors, who had to work with squares and rectangles as they surveyed pieces of land for building purposes. The procedures ultimately worked out were preserved by the scribes on clay tablets, hundreds of which have been discovered by archaeologists in the twentieth century. These tablets, written in cuneiform writing, are now preserved in numerous libraries around the world. They have been translated and extensively studied. Although the tablets generally just present problems and procedures for their solutions, scholars have determined that the procedures were worked out by methods similar to what is presented here. Activity 1: The width of a rectangle is 10 units. Its length is unknown. We place a square on one of the sides of the rectangle representing the length as shown in the figure. Together, the two shapes have an area of 39 square units. What is the length of the rectangle?

Hint: Begin by slicing the rectangle with sides 10 and x in half vertically and moving one half to the bottom of the square.

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Next, try the same problem with other numerical values. 1. 2. 3. 4.

Width = 8; Total area = 105 Width = 14; Total area = 120 Width = 5; Total area = 14 Width = 7; Total area = 78

Work out in your group a verbal description of the method you have decided to use. Make sure you describe the steps very explicitly. Activity 2: Make up and solve five of your own similar problems for each of the specific conditions listed here: a. b. c.

Both sides of the rectangle must be whole numbers. Both sides of the rectangle must be rational numbers, with at least one side not a whole number. The length of your rectangle must be an irrational number..

Once you have solved your problems, exchange your problems with another group and solve those of that other group. Activity 3: It is now time to do this problem generally, without using concrete numbers for the width or the area. In fact, we want to construct a formula that will enable you to find the answers to all the problems you solved in the previous activities. Thus, we will use letters to represent the numbers from those problems. We will use b to represent the given width of the rectangle and c to represent the total area. And, as usual in algebra, we will use x to represent the unknown length of the rectangle. Thus, our general problem is the following: Given a rectangle with width equal to b. We place a square on the side representing the length. The total area of the square and the rectangle is equal to c. What is the unknown length of the rectangle? b

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We first want to translate the problem into an algebraic equation and then use the procedures you have developed in activities 1 and 2 to develop a formula for finding the unknown. The area of a square of side x is _______________. The area of a rectangle with width b and length x is ______________. The equation stating that the sum of the two previous areas is equal to c is _________________ Now, in your group, translate the procedure you developed to solve the problems in activities 1 and 2 into a formula. You should get a result equivalent to the following: 2

b ⎛b⎞ x = ⎜ ⎟ +c − . 2 ⎝2⎠ That is, this formula gives you a solution to the quadratic equation x2 + bx = c. Now, rewrite each of the problems from activities 1 and 2 in the form of a quadratic equation with concrete numbers instead of the letters b and c. Then use the formula to determine the value of x. The answers should be the same as those you found previously. Activity 4: We want to generalize the formula found above so it applies to other quadratic equations. First, we want to apply it to quadratic equations of the form ax2 + bx = c. To do this, divide both sides of the equation by a. The new equation is then _______________________. Now, apply the formula to this equation. That is, replace b in the previous formula by the new coefficient of x, namely, ______, and c in the previous formula by the new constant term, namely, ______. The new formula is then ______________________________________ You should note that in the formula, it does not matter whether a, b, or c is positive or negative. In ancient Mesopotamia, people did not consider negative numbers at all, but we can and should consider them. And even though it is not obvious how to interpret the squares and rectangles above when the coefficients are negative, it turns out that the formula works in any case. Thus, we will now apply the formula to solve the most general quadratic equation: ax2 + bx + c = 0. We need to rewrite this in the form for the formula we just derived, that is, with the constant term on the right side of the equation. So this becomes _________________________. The difference between this and the equation we used before is simply that c has been replaced by –c. So, making this replacement in the formula you derived above, we get the new formula ______________________________

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There are two modifications we usually make to give a more standard version of the formula. First, we use algebra to simplify the expression inside the radical sign. That is, instead of having 2

⎛ b ⎞ ⎜ ⎟ − c , we replace this with ⎝ 2a ⎠

b 2 − 4 ac . Make sure you understand how to do this. 2a

Second, given that if the expression inside the square root sign is positive, there are two square roots, a positive and a negative value, we include both in the quadratic formula. We also put the term –b/2a first before the term with the plus or minus. We therefore will write the formula in one of the two following forms: x=−

b b 2 − 4 ac ± 2a 2a

or x =

−b ± b 2 − 4ac . 2a

Exercises: Solve each of the following quadratic equations using the quadratic formula in either of the two forms presented last. 1.

x2 + 15 = 8x

x2 + 3 = 4x

x2 – 11x + 24 = 0

x2 – 10x + 5 = 0 (Note that the answers here are irrational numbers; just express them in terms of radicals.)

x2 + 7 = 9x

x2 – 6x + 8 = 0

x2 + 2x + 6 = 0 (What is happening here? Why? Can you generalize this?)

x2 + 4x = 45

x2 + 8x + 9 = 0

10. x2 + 5x – 24 = 0 11. x2 + 6x + 18 = 0 12. x2 + 4x = 3 13. 2x2 + 5x + 3 = 0 14. 3x2 + 8x + 5 = 0 15. 3x2 + 13x + 4 = 0

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Problems Involving Quadratic Equations Teacher Notes Level: This activity is designed for use in the second year algebra course in high school. Materials: Student pages Time Frame: Two class sessions, with homework between them. Objectives: To understand the types of problems a knowledge of quadratic equations will help us solve. When to Use: This activity can be used after the students have learned how to solve quadratic equations using the quadratic formula. How to Use: This material is designed for independent work or work in small groups. Whole class discussions will be useful after students have tried to solve some of the problems. Answers to Exercises: 1. 12

10. 6, 19 12

2. 3, 7

11. 13, 9

3. 6, 4

12. 1 +

4. 9

13. Yes, twice: 1 ±

5. 24

14. Yes, 1 m.

√ √ 6. 5 5 − 5, 15 − 5 5

15. No

7. 4, 6

16. 7

q √ √ 8. 14 − 1 11 − 2 13 − 2 11

17. 2 12 , 10

√ 26 m. 1 2

√ 2 m.

9. 3 13 , 3

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Problems Involving Quadratic Equations Student Pages Historical Background Recall that the first algebra text was written by Mohammed ibn Musa al-Khwarizmi in Baghdad, about 825 C.E. Interestingly, al-Khwarizmi wrote in the preface to his work that it was written to demonstrate what is “easiest and most useful in arithmetic, such as men constantly require in cases of inheritance, legacies, partition, law-suits, and trade, and in all their dealing with one another, or where the measuring of lands, the digging of canals, geometrical computation, and other objects of various sorts and kinds are concerned.” In other words, al-Khwarizmi claimed he was writing a practical manual, not a theoretical one. But what were the practical uses of quadratic equations according to al-Khwarizmi? A search of the many problems he put in his book actually finds very few. Other Islamic authors over the years also wrote texts demonstrating how to solve quadratic equations, but “practical” problems were few and far between. In the Renaissance, once algebra reached Europe, numerous authors wrote texts expounding ideas on the solution of quadratic equations. But again, there were few problems other than purely abstract ones. We will, however, look at some examples of practical problems in the algebra texts of Gerolamo Cardano (1545) and Robert Recorde (1557).

Quadratic Problems 1. Here is al-Khwarizmi’s first problem: I have divided ten into two parts; I have multiplied the one of the two parts by the other; after this, I have multiplied the one of the two by itself, and the product of the multiplication by itself is four times as much as that of one of the parts by the other. Let us follow al-Khwarizmi’s solution as well, although we will use symbols where he just used words. We suppose one of the parts of 10 is x. Then the other is 10 − x. We multiply one of the two parts by the other; that is, we multiply x by 10 − x and get 10x − x2 . We then multiply one part by itself, so we multiply x by x to get x2 . The problem states that this product is four times as much as the previous product. Therefore, we get the equation x2 = 4(10x − 4x2 ) or x2 = 40x − 4x2 . To solve this quadratic equation, we begin by adding 4x2 to each side. We get 5x2 = 40x. In this case, we do not need to use our formula. We just note that if we divide both sides by x, we can reduce this equation to 5x = 40, an equation whose solution is x = 8. Thus al-Khwarizmi’s solution to the problem is that the two parts of 10 are 8 and 2. But there is a slight difficulty with his procedure. Namely, we divided both sides of an equation by x. We can only do this if we assume that x is not 0. In this case, if x is 0, we have a perfectly good solution to the equation, since certainly 5 · 02 = 40 · 0. So for us, if not for al-Khwarizmi, there are two solutions to the original problem. One is the solution 8 and 2, while the other is the solution 0 and 10. You may argue, of course, that in this second solution we have not “divided” 10 into two parts. So perhaps we c 2004, M.A.A.

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should reject this second solution after all and stay with al-Khwarizmi’s one solution. 2. Let us try another problem from al-Khwarizmi: I have divided ten into two parts; I have then multiplied each of them by itself, and when I added the products together, the sum was 58. To solve this problem we begin as before by taking the two parts of 10 to be x and 10 − x. The squares of these two parts of then x2 and (10 − x)2 = 100 − 20x + x2 . We then add these two squares together and set the sum equal to 58: x2 + 100 − 20x + x2 = 58. This simplifies to 2x2 − 20x + 42 = 0, or, if we divide every coefficient by 2, to x2 − 10x + 21 = 0. Now we can use the quadratic formula to get −10 ± x=− 2

(−10)2 − 4 · 1 · 21 2

√ √ 4 100 − 84 16 =5± =5± = 5 ± = 5 ± 2 = 7, 3. 2 2 2

The two numbers into which we divide 10 are then 7 and 3. We can check that the sum of the squares of these two numbers are 49 + 9 = 58, as required. 3. Here is a somewhat more complicated problem of al-Khwarizmi: I have divided ten into two parts; and have divided the first by the second, and the second by the first. The sum of the quotients is 2 16 . Again, we let the two parts of 10 be x and 10 − x. The equation is then 10 − x x 1 + =2 . x 10 − x 6 This equation does not at first glance appear to be a quadratic equation, because there are no squares. But to simplify it, we must multiply each term by the common denominator, and then squares will appear. The common denominator of the three terms is 6x(10 − x), so we multiply each term by that quantity. The equation then becomes 1 6(10 − x)(10 − x) + 6x · x = 6 · 2 · x(10 − x), 6 which simplifies to 6(10 − x)2 + 6x2 = 13x(10 − x) or to 600 − 120x + 6x2 + 6x2 = 130x − 13x2 . If we move the terms on the right to the left and collect common terms, we get 25x2 − 250x + 600 = 0, or, dividing everything by 25, x2 − 10x + 24 = 0. The solution is then −10 x=− ± 2 c 2004, M.A.A.

(−10)2 − 4 · 1 · 24 2

√ √ 2 100 − 96 4 =5± =5± = 5 ± = 5 ± 1 = 6, 4. 2 2 2 40

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Again, there are two answers, with one of them representing one part of 10 and the other the second part. That is, if we divide 10 into the two parts 6 and 4, then 64 + 46 = 2 16 , as demanded. 4. Problems 1–3 are typical of the problems al-Khwarizmi puts into his book. And it is clear that these are not “practical” problems, as promised in his preface. But he does have a few problems which might appear to be practical. Here is one of them: One dollar is to be divided equally among a certain number of men. If you add one man more to them and divide again, the quota of each is one-sixth of a dollar less than the first time. How many men were there originally? (The value of a dollar was more in those days then it is today!) To solve this problem, we let x be the original number of men. Then each man will receive x1 dollars. If we add 1 more man, we will have x + 1 men and each man will 1 receive x+1 dollars. The problem states that the first quotient is 16 more than the second. Therefore, the equation is 1 1 1 − = . x x+1 6 To solve this, we proceed as in the previous problem and multiply all the terms by the common denominator, 6x(x + 1). We then get the new equation 6(x + 1) − 6x = x(x + 1)

6x + 6 − 6x = x2 + x.

We can simplify this equation to x2 + x − 6 = 0, an equation we can again solve by use of our formula. We get 1 x=− ± 2

12 − 4 · 1 · (−6) 2

1 =− ± 2

√

√ 1 1 5 1 + 24 25 =− ± = − ± = 2, −3. 2 2 2 2 2

In this case, one of our answers is negative, and this cannot be a solution to the problem of finding the number of men. Therefore, the correct solution is x = 2, and there were originally 2 men. We note that in that case, each man would receive 12 of a dollar, and this is 16 more than the 13 of a dollar that 3 men would receive. You may not think that this problem is very practical, but that seems to be the best that al-Khwarizmi could figure out as to a practical use of quadratic equations. 5. Cardano has a few more practical-sounding problems which can be solved by the use of quadratic equations. Here is one problem of Cardano: There were two associations, one of which had three more members than the other. They divided equal numbers of gold pieces among their members. The number of gold pieces to be divided in each case was 93 more than the total of the members in the two associations, and the members of the smaller association each received 6 more gold pieces than the members of the larger association. What was the membership of each association? If we let x be the number of members in the smaller association, then the number of members in the larger association is x + 3. So the total membership of the two associations is x + x + 3 = 2x + 3, and therefore the number of gold pieces to be divided in c 2004, M.A.A.

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each case is 2x + 3 + 93 = 2x + 96. Our equation is then 2x + 96 2x + 96 = + 6, x x+3 since each member of the smaller association receives 6 more gold pieces than does each member of the larger association. To solve this equation, we multiply each term by the common denominator, x(x + 3). We then get (2x + 96)(x + 3) = 6x(x + 3) + (2x + 96)x. We can rewrite this equation as 2x2 + 102x + 288 = 6x2 + 18x + 2x2 + 96x. This simplifies to 6x2 + 12x − 288 = 0, or, dividing by 6, to x2 + 2x − 48 = 0. You can solve this by the quadratic formula to get two answers: x = 6 and x = −8. Since the negative answer makes no sense in this problem, we know that the number of members in the smaller association is 6 and therefore that the number in the larger association is 9. The number of gold pieces in each case is 108. 6. Here is one more problem from Cardano’s book: A certain man went to market three times. The first time he brought back twice what he had taken with him. On the second trip, he took with him this double amount and returned with the same plus its square root plus 2 additional gold pieces. All this he preserved, and he returned to market with it a third time. His profit from this trip was the square of what he took with him and 4 gold pieces in addition. He returned, moreover, with 310 gold pieces. How many gold pieces did he take with him on his first trip? We solve this problem in stages. First, let x be the number of gold pieces he took with him on the third trip. He returned with that amount plus his profit, that is, with x + x2 + 4 gold pieces, and this was equal to 310. So we have the equation x2 + x + 4 = 310, or x2 + x − 306 = 0. We can solve this to get q

√ √ 12 − 4 · 1 · (−306) 1 1 1 35 1 + 1224 1225 1 =− ± =− ± =− ± = 17, −18. x=− ± 2 2 2 2 2 2 2 2 We reject the negative answer, and find that the man took 17 gold pieces with him on his third trip. c 2004, M.A.A.

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Now, let y 2 be the amount he took with him on the second trip. We use y 2 , because we need to take its square root, and this lets us do it easily. That is, he returned from the second trip with y 2 plus y, its square root, plus 2, and this is equal to the amount he began the third trip with, namely 17. So our equation is y 2 + y + 2 = 17

y 2 + y − 15 = 0.

We solve this equation again using the formula. We get 1 y=− ± 2

12 − 4 · 1 · (−15) 2

1 =− ± 2

√ √ 1 1 + 60 61 =− ± . 2 2 2

In this case, we cannot simplify the square root, so we will just leave it.√ Of course, we must reject the negative result, as before, so our solution is y = − 12 + 261 . But recall that the amount he took with him on the second trip was y 2 . So we need to square this value. We get √ !2 √ √ 61 62 1 1√ 1 1 61 61 61 + = − = 15 − = − 61. y2 = − + 2 2 4 2 4 4 2 2 2 Finally, since the man brought with him on the first trip half of what he brought with him on the second trip, we find that the amount he took with him initially is half of y 2 , or 3 1√ 61. 7 − 4 4 7. We now look at one more problem from a Renaissance text, one which is almost practical and seems to be the earliest quadratic problem to deal with a question of velocity and distance. This problem is from The Whetstone of Witte, published in 1557 by the first English author of mathematical works in the Renaissance, Robert Recorde: There is a strange journey appointed to a man. The first day he must go 1 12 miles, and every day after the first he must increase his journey by 16 of a mile, so that his journey shall proceed by an arithmetical progression. And he has to travel for his whole journey 2955 miles. In what number of days will he end his journey? To solve this problem, we need to know how to find the sum of an arithmetical progression, or what we tend to call it today, an arithmetic sequence. Such a sequence is a sequence which starts with a particular number a. The next terms in the sequence are a + d, a + 2d, a + 3d, and so on, where d denotes the common difference between successive terms. If there are n terms in the sequence, then the sequence is of the form a + (a + d) + (a + 2d) + (a + 3d) + · · · + (a + (n − 2)d) + (a + (n − 1)d). Recorde notes that the sum of the first and last terms is 2a + (n − 1)d and that this is also the sum of the second and second to last terms, the third and third to last terms, and so on. Therefore, if there are an even number of terms, we can find the sum by multiplying the sum of the first and last terms by n2 , half the number of terms. This c 2004, M.A.A.

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also works if there are an odd number of terms, for in that case, the middle term is half the sum of the first and last term. Then the sum would be found by taking n−1 2 times the sum of the first and last terms and adding to this half the same sum. This amounts to the same thing as multiplying the sum of the first and last terms by n2 , half the number of terms. To solve his problem, then, Recorde needs to calculate the sum of the arithmetic sequence whose first term in a = 32 = 96 and whose difference is d = 16 . If there are n terms in the sequence, the last term is 96 + (n − 1) 16 = 9+n−1 = n+8 6 6 . It follows that the n+17 sum of the first and last terms is 6 and therefore that the sum of the sequence is S=

n2 + 17n n n + 17 · = . 2 6 12

Here n is the number of days of the journey, and therefore the equation we need to solve is n2 + 17n = 2955 or n2 + 17n = 35, 460. 12 If we rewrite this as n2 + 17n − 35, 460 = 0 and apply the quadratic formula, we get q

√ 172 + 4 · 35, 460 17 17 289 + 141, 840 =− ± n=− ± 2 2 2 2 √ 17 17 377 142, 129 =− ± =− ± = 180, −197. 2 2 2 2 As before, we reject the negative answer and get that the journey will take 180 days. Exercises: The first six problems are taken from the work of al-Khwarizmi. 1. I have multiplied one-third of a quantity plus one by one-fourth of the same quantity plus one, and the product is twenty. What is the quantity? 2. I have divided ten into two parts, and multiplying one of these parts by the other, the result is twenty-one. What are the parts? 3. I have divided ten into two parts, and having multiplied each part by itself, I have added them together, and have added to their sum the difference of the two parts, and the total is fifty-four. What are the parts? 4. Find a number which, if one-third of it be added to 3, and the sum be subtracted from the number, and the result multiplied by itself, the answer is the original number. 5. If you begin with a number, then subtract one-third of it, then one-fourth of it, and then four, and then if you multiply the result by itself, you get twelve more than the original number, find the number. c 2004, M.A.A.

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6. I have divided ten into two parts; I have multiplied one part by ten and the other by itself, and the products were the same. What are the parts? The next two problems are taken from the work of Cardano. 7. There were two leaders each of whom divided 48 gold pieces among his soldiers. One of these had two more soldiers than the other. The one who had two soldiers fewer had 4 gold pieces more than the other for each soldier. How many soldiers did each leader have? 8. There is a number which, if twice its square root be added to it and if to this sum twice its square root is added, gives a total of 10. Find the number. (Hint: Do this in two steps, beginning with the final sum; let this second sum be x2 , so you can easily find its square root. The square of the solution will then be the first sum.) The next three problems are taken from the work of Recorde. 9. There are two men that have silk to sell. One has 40 yards and the other 90. The first man’s silk is not so fine as the second man’s silk, so he sells 13 of a yard more for every dollar than the second man does. At the end, both of them have together received 42 dollars. How much did each sell for a dollar? 10. There were two men who had certain sums of money, such that the second man’s sum was 3 14 times the first. If their two quantities were multiplied together and to that total the two quantities were added, there would amount 142 12 . How much did each have? 11. There is a number, which I have forgotten, and it is divided into 2 parts, whereof the first I have forgotten also. But the second was 4, and yet this I remember, that if the part which I have forgotten be multiplied by itself and then also with 4, those 2 numbers will add up to 117. I would like to know what was the original number, and also what is the part which I have forgotten? We now look at some more modern problems: 12. Under certain conditions, the path of a diver diving off of a cliff is given by y = −x2 + 2x + 25, where x is the distance of the diver from the cliff and y is the height of the diver above the water, where both x and y are measured in meters. How far from the cliff will the diver be when she enters the water, that is, when y = 0? 13, Will the diver in exercise 12 ever reach a height of 25.5 meters? How far from the cliff will the diver be at this point? 14. Will the diver in exercise 12 ever reach a height of 26 meters? How far from the cliff will the diver be at this point? 15. Will the diver in exercise 12 ever reach a height of 27 meters? How far from the cliff will the diver be at this point? c 2004, M.A.A.

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16. If a rock is tossed upwards at a velocity of 96 feet per second from a cliff which is 112 feet above ground level, then the height h in feet of the rock above the ground t seconds after it is thrown is given by the formula h = â&#x2C6;&#x2019;16t2 + 96t + 112. At what time will the rock hit the ground, that is, at what time will h = 0? 17. Suppose the height h in meters of a ball t seconds after it is hit on the moon is described by the formula h = â&#x2C6;&#x2019;0.8t2 + 10t. At what times will the ball be at height 20 meters?

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Descartesâ&#x20AC;&#x2122;s Method for Solving Quadratic Equations Geometrically Teacher Notes Level: This activity is designed for use in the second year algebra course in high school. Materials: Student pages Time Frame: 15-30 minutes. Objectives: To understand that solutions to quadratic equations may be geometric as well as algebraic or numerical. To practice geometrical ideas. When to Use: This activity can be used after the students have learned how to solve quadratic equations using the quadratic formula. It is also assumed that they have a basic knowledge of geometrical constructions. How to Use: The material on the student pages can be presented by the teacher. The exercises then could be done in small groups with a whole class discussion afterwards.

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Descartes’s Method for Solving Quadratic Equations Geometrically Student Pages Descartes introduced his work, the Geometry, as follows: “Any problem in geometry can easily be reduced to such terms that a knowledge of the lengths of certain straight lines is sufficient for its construction.” His goal then was to represent lengths by letters and manipulate the letters algebraically in order to solve geometric problems. One of the problems with which he began was to find the solution of the quadratic equation x2 = bx + c2 , where b and c are given positive numbers. (Descartes used the term c2 in his equation rather than simply using c, because he was thinking of this equation in geometric terms. Since the left side was a square, the right side had to represent an area. One can think of bx as representing the area of a rectangle. Thus the remaining term had to represent an area as well. For mathematicians before Descartes, the term c would be thought of as a line segment.) Descartes began his solution by constructing a right triangle NLM with LM = c and LN = 12 b

He then extended the hypotenuse to O, where NO = NL and constructed the cir√ 2 + LM 2 = NL cle centered on N with radius NO. Because ON = b/2 and NM = q q (b/2)2 + c2 + b2 /4 + c2 , he concluded that OM = ON + NM is the required value x. After all, the value of x is given by the standard formula 1 x = b+ 2

1 2 b + c2 , 4

and this is equal to the length OM.

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Exercises: 1. Using the same diagram, show that MP is the solution to x2 = −bx + c2 . 2. Modify the diagram by drawing MQR parallel to LM , intersecting the circle at the two points Q and R. Show that MQ and MR are the two solutions to x2 = bx − c2 .

3. In this discussion, Descartes limited himself to positive solutions. Although he wrote coefficients in his equations which were positive numbers preceded by minus signs, he still felt constrained, like al-Khwarizmi, to write down three separate types of quadratic equations. Nevertheless, the equations x2 = bx + c2 and x2 = −bx + c2 will, in general, have a negative solution as well as the positive ones found above. How can you represent these negative solutions using the diagram? 4. A quadratic equation of the form x2 = bx−c2 may not have the two solutions indicated in exercise 2. It may have only one solution, or no real solutions at all. How would these two circumstances show up on your diagram?

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Finding Cube Roots Teacher Notes Level: Second year high school algebra course. Materials: Student pages; calculator Time Frame: Each part of this activity can be done in one class period. Objectives: To learn the binomial theorem for n = 3. To demonstrate hand methods of calculation of cube roots. When to Use: The basic prerequisite for calculating cube roots is a knowledge of the binomial theorem for n = 3. This is covered in the first part of the activity. Finding cube roots should be done before treating the solution of cubic equations in general. How to Use: Explain to the class that you will be considering methods for solving the simple cubic equation x3 = √ N , for N a positive number. Although one could express the solution to this equation as 3 N , what we want to do here is find a numerical approximation to that solution. After working through the derivation of the binomial theorem with the entire class, it may be useful to assign the two calculation methods to different groups and have the groups compare results afterwards. Answers to Exercises: Binomial Theorem

Chinese Method

1. x3 + 9x2 + 27x + 27

1. 123

2. 8x3 + 36x2 + 54x + 27

2. 14,300

3. 8 + 12x2 + 6x4 + x6

3. 12 12

4. 125x3 − 225x2 + 135x − 27

4. 39 78

5. 27x6 − 54x4 + 36x2 − 8

5. 124 23

6. x6 + 9x4 y + 27x2 y 2 + 27y 3

Indian Method

7. 8x9 + 60x6 y 2 + 150x3 y 4 + 125y 6

2. 37

9. a5 + 5a4 b + a0a3 b2 + a0a2 b3 + 5ab4 + b5

3. 237

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Finding Cube Roots Student Pages The Binomial Theorem Recall that the binomial theorem for a square states that (a + b)2 = a2 + 2ab + b2 . To extend this to cubes, we multiply each side of this equation by a + b, expand the right side by the distributive law, and simplify. We get (a + b)3 = (a + b)(a2 + 2ab + b2 ) = a3 + 2a2 b + ab2 + a2 b + 2ab2 + b3 = a3 + 3a2 b + 3ab2 + b3 . Thus the binomial theorem for cubes states that (a + b)3 = a3 + 3a2 b + 3ab2 + b3 We can state this result geometrically as follows: If a quantity is divided into two parts, the cube of the whole is equal to the cubes of the two parts plus three times the solid whose height is the first part and whose base is the square of the second part plus three times the solid whose height is the second part and whose base is the square of the first part. (See below for a diagram.) Examples: 1. To expand (a + 2)3 , we use the basic formula: (a + 2)3 = a3 + 3a2 · 2 + 3a · 22 + 23 = a3 + 6a2 + 12a + 8. 2. To expand (3x + 4)3 , we use the formula, replacing a by 3x and b by 4: (3x + 4)3 = (3x)3 + 3(3x)2 · 4 + 3(3x) · 42 + 43 = 27x3 + 3 · 9x2 · 4 + 9x · 16 + 64 = 27x3 + 108x2 + 144x + 64. 3. To expand (1 + x)3 , we again use the formula, taking a = 1 and b = x. We get (1 + x)3 = 13 + 3 · 12 · x + 3 · 1 · x2 + x3 = 1 + 3x + 3x2 + x3 . 4. We can still use the formula, even if we have a minus sign. To expand, for example, (2x − 1)3 , we write (2x − 1)3 = (2x)3 + 3(2x)2 (−1) + 3(2x)(−1)2 + (−1)3 = 8x3 + 3 · 4x2 (−1) + 6x · 1 + (−1) = 8x3 − 12x2 + 6x − 1. c 2004, M.A.A.

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5. Here is one more example, with both a minus sign and power of x: Expand (x2 − 2)3 . We have (x2 − 2)3 = (x2 )3 + 3(x2 )2 (−2) + 3(x2 )(−2)2 + (−2)3 = x6 − 6x4 + 12x2 − 8. Exercises: Expand each of the algebraic expressions in problems 1 through 7 by use of the binomial theorem for cubes: 1. (x + 3)3 2. (2x + 3)3 3. (2 + x2 )3 4. (5x − 3)3 5. (3x2 − 2)3 6. (x2 + 3y)3 7. (2x3 + 5y 2 )3 8. By following the procedure by which we derived the binomial formula for cubes from the binomial formula for squares, derive the binomial formula for fourth powers. That is, show that (a + b)4 = a4 + 4a3 b + 6a2 b2 + 4ab3 + b4 . 9. Using the result of exercise 8, discover the binomial theorem for fifth powers. That is, discover the expansion formula for (a + b)5 .

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Chinese Cube Root Method The classic Chinese work, the Nine Chapters on the Mathematical Art, written near the beginning of our era, has several problems dealing with determining the cube root of a number. The idea behind the method described there comes from using the binomial expansion (r + s)3 = r3 + 3r2 s + 3rs2 + s3 . This expansion represents the breaking up of a cube of side r + s into eight pieces, one a cube of side r, one a cube of side s, three a box with two sides r and one side s, and three a box with one side r and two sides s.

Keep the diagram in mind as we follow an example. Consider the solution of the equation x3 = 12, 812, 904. We first note that the answer (or at least the integral part of the answer) will be a three digit number (why?), where the first digit is 2 (why?) In other words, the closest integer solution can be written as x = 200 + 10b + c. We determine b, by first ignoring the value c. That is, we think of the cube as having side 200 + 10b and volume 12,812,204. We therefore want to find the largest b so that (200 + 10b)3 = 2003 + 3 · 2002 · 10b + 3 · 200 · (10b)2 + (10b)3 ≤ 12, 812, 904. Subtracting off the cube of 200 (or 8,000,000) from both sides, we need b to satisfy the inequality 3 · 2002 · 10b + 3 · 200 · 100b2 + 1000b3 ≤ 4, 812, 904. Geometrically, the left side of this inequality is the volume of the seven pieces of the box from our diagram, leaving off the cube of side 200. We rewrite the left side of this inequality as b(1, 200, 000 + 60, 000b + 1000b2 ) and try to find the largest b so that this expression is less than or equal to 4,812,904. We try in turn b = 1, 2, 3, . . ., and eventually discover that b = 3 is the largest value satisfying the inequality. We therefore replace b by 3 in the expression b(1, 200, 000 + 60, 000b + 1000b2 ) to get 3(1, 200, 000 + 60, 000 · 3 + 1000 · 32 ) = 4, 167, 000. Our next step is to subtract 4,167,000 from 4,812,904, leaving 645,904. We can also think of this as subtracting 2303 from the original value 12,812,904. So, returning to our diagram, we now think of the cube as having side 230 + c with volume 12,812,904, from which we have subtracted the cube of side 230. We therefore need an inequality for the c, and that is determined by finding the volume of the seven pieces we have left. Namely, we need c to satisfy 3c·2302 +3c2 ·230+c3 ≤ 645, 904, or, equivalently, c(3·2302 +3·230c+c2 ) ≤ c 2004, M.A.A.

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645, 904. Again, we try different values of c to find the largest satisfying the inequality. In this case it turns out that c = 4 satisfies this as an equality, so the solution to the original equation is x = 234. Our answer is an integer. If it were not, we could continue the process using decimals, where at each further step we add on three zeroes after the decimal point. One obvious question is how do we â&#x20AC;&#x153;guessâ&#x20AC;? the appropriate digits without doing a lot of arithmetic. One way to do this is simply to ignore the powers of b (or c) larger than the first, and try to determine a digit which satisfies the linear inequality. Sometimes, the solution to the linear inequality will be the value needed. But if it is not, the next lower digit will generally work. Exercises: Use the Chinese procedure to find the cube roots of the following numbers. These examples are all taken from the Nine Chapters. Since some of these involve fractions, you may need to figure out a way to modify the method slightly at the end to take care of these. 1. 1,860,867 2. 2,924,207,000,000 3. 1953 18 4. 63, 401 447 512 5. 1, 937, 541 12 27

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Indian Cube Root Method Aryabhata (b. 476), the earliest Indian mathematician whose name is known, included an algorithm for determining cube roots in his book, the Aryabhatiya, written in 499. Little is known about the author, other than he lived at Kusumapura, near the Patalipura (modern Patna) on the Ganges in Bihar in northern India. This city was the capital city of the Gupta dynasty, the dynasty that ruled northern India in the fourth and fifth centuries. The Aryabhatiya is a brief book in Sanskrit of four sections and 123 stanzas, the second section of 33 stanzas dealing with mathematics. By no means a detailed working manual, it was only a brief descriptive work perhaps intended to be memorized and surely only intended as a summary of either a more detailed treatise or simply of lectures given by the author. Stanza II, 5 of this work provides a rule for calculating cube roots. We present a fairly literal translation of the original, and then give a more detailed explanation with an example. One should divide the second aghana by three times the square of the cube root of the preceding ghana. The square (of the quotient) multiplied by three times the purva (that part of the cube root already found) is to be subtracted from the first aghana and the cube (of the quotient of the above division) is to be subtracted from the ghana. The technical terms in the stanza refer to the places in the given number. Counting from right to left, the first, fourth, and so on places are named ghana (cubic), the second, fifth, and so on are called the first aghana (non-cubic), while the third, sixth, and so on are called the second aghana. But as the example of calculating the cube root of 12,977,875 shows, certain steps are not spelled out explicitly, probably due to the limitations of Sanskrit verse. Note that this algorithm is strictly arithmetical, but that, like the Chinese algorithm, is based on the binomial expansion for n = 3. As in that algorithm, the process begins by determining the number of digits in the answer as well as the first digit.

1587

12977875 _8 | 49 36 137 54 837 27 |8 1 0 8 7935 1737 1725 125 125

)2 )3

First digit 2 ≈ 3 12 23 12 = 3 × 22 3 = quotient (4 is too large) 32 multiplied by 3 × 2

33 1587 = 3 × 232 5 = new quotient 52 multiplied by 3 × 23 53

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Exercises: 1. Show how the binomial expansion for n = 3 is involved in this algorithm. You will need to check where the coefficients 1, 3, 3, 1 of that expansion show up here. 2. Calculate the cube root of 50,653 using this algorithm 3. Calculate the cube root of 13,312,053 using this algorithm.

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Solving Cubic Equations via Conic Sections Teacher Notes Level: This activity is designed for use in a precalculus course after students have studied the analytic geometry of conic sections. Materials: Historical background; student pages; graphing calculator; graph paper Time Frame: One class period, with homework, should be sufficient. Objectives: To extend studentsâ&#x20AC;&#x2122; knowledge of solving equations by showing that there are geometric, as well as algebraic and numerical, methods for solving cubic equations. To further develop studentsâ&#x20AC;&#x2122; understanding of conic sections. When to Use: This activity can be used after the students have been introduced to the analytic geometry of conic sections. How to Use: This material can be covered in class by the teacher. In this case, the student pages will serve essentially as textbook material to be read before and after the class presentation. But the material could also be assigned for independent work or small group work. In either case, the historical background should be assigned as independent reading. If the material is assigned for small group work, there could be a whole class discussion of different possibilities for solving given cubic equations via conic sections.

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Solving Cubic Equations via Conic Sections Student Pages Historical Background By the end of the ninth century Islamic mathematicians, having read the major Greek texts, had noticed that certain geometric problems led to cubic equations, equations which could be solved through finding the intersection of two conic sections. One of these problems was the classical problem of doubling the cube, that is, of constructing geometrically a cube of volume double another one. If the given cube had side a, this problem amounted to constructing a length x such that x3 = 2a3 , that is, to solving a cubic equation. Ancient accounts recorded that Hippocrates of Chios (mid-fifth century BCE) had reduced this problem to finding two mean proportionals x, y, between a and 2a. In other words, if a : x = x : y = y : 2a, then a3 : x3 = (a : x)3 = (a : x)(x : y)(y : 2a) = a : 2a = 1 : 2 and therefore x3 = 2a3 , as desired. Hippocrates, however, could not figure out how to construct these two mean proportionals. It was probably Menaechmus (fourth century BCE) who first constructed curves from which the two mean proportionals could be found. In modern terms, solving the proportions a : x = x : y = y : 2a is equivalent to solving simultaneously any two of the three equations x2 = ay, y 2 = 2ax, and xy = 2a2 , equations that represent parabolas in the first two instances and a hyperbola in the third. Thus, if one could construct these curves, the coordinates (x, y) of the intersection point would be the two mean proportionals necessary to solve the cube doubling problem. Although the Greeks solved a few other cubic equations similarly, they never tried to attack the general problem of cubics systematically. During the tenth and eleventh centuries, Islamic mathematicians studied these Greek problems. It was the mathematician and poet ‘Umar ibn Ibr¯ah¯ım al-Khayy¯am¯ı (1048–1131) (usually known in the West as Omar Khayyam), who was finally able systematically to classify and then solve all types of cubic equations by this general method of intersecting carefully chosen conic sections. AlKhayy¯am¯ı’s major mathematics text, the Treatise on demonstrations of problems of al-jabr and al-muqabala, is primarily devoted to the solution of cubic equations. As the author makes clear in the preface, the reader of the work must be thoroughly familiar with Euclid’s Elements and the first two books of Apollonius’s Conics. The latter work discussed properties of conic sections in great detail, and these properties were necessary for Omar’s solution. Nevertheless, al-Khayy¯am¯ı’s text addressed algebraic, not geometric, problems, and the author would have liked to have provided algebraic algorithms for solving cubic equation, analogous to al-Khw¯arizm¯ı’s three algorithms for solving quadratic equations. As he wrote, “When, however, the object of the problem is an absolute number, neither we, nor any of those who are concerned with algebra, have been able to solve this equation — perhaps others who follow us will be able to fill the gap.” As we will see in another section of this module, the gap was finally filled in sixteenth century Italy. c 2004, M.A.A.

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Using Conic Sections to Solve Cubic Equations Al-Khayy¯am¯ı began his work by giving a complete classification of equations of degree up to three. Since for al-Khayy¯am¯ı, as for his predecessors, all numbers were positive, he had to list separately the various forms that might possess positive roots. Among these were fourteen types of cubic equations not reducible to quadratic or linear equations. These were in three groups, one binomial equation, x3 = d, six trinomial equations, x3 + cx = d, x3 + d = cx, x3 = cx + d, x3 + bx2 = d, x3 + d = bx2, and x3 = bx2 + d; and seven tetranomial equations, x3 + bx2 + cx = d, x3 + bx2 + d = cx, x3 + cx + d = bx2 , x3 = bx2 + cx + d, x3 + bx2 = cx + d, x3 + cx = bx2 + d, and x3 + d = bx2 + cx. (Why does the equation x3 + bx2 + cx + d = 0 not appear?) Each of these equations was analyzed in detail by the author. In particular, he found and completely described two conic sections whose intersection provides at least one solution to the equation, proved that his solution was correct, and finally discussed the conditions under which there may be no solutions or more than one solution. We will discuss here al-Khayy¯am¯ı’s solution of x3 + cx = d or, as he puts it, the case where “a cube and sides are equal to a number.” To construct the solution, al-Khayy¯ am¯ı √ set AB equal in length to a side of the square whose area is c; that is, AB = c (See the figure.) He then constructed BC perpendicular to AB so that BC ·AB 2 = d, or BC = d/c. Next, he extended AB in the direction of Z and constructed a parabola with vertex B, axis BZ, and parameter AB. (The “parameter” is the constant p in the general equation of a parabola x2 = py.) In modern notation, al-Khayy¯am¯ı’s parabola has the equation √ 2 x = cy. Similarly, he constructed a semicircle on the line BC. Its equation is d x− 2c

+y =

d 2c

d − x = y2. x c

The circle and the parabola intersect at a point D. It is the x coordinate of this point, here represented by the line segment BE, which provides the solution to the equation.

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Al-Khayy¯am¯ı proved that his solution is correct by using the basic geometric properties of the parabola and the circle. We will prove it using algebra. Namely, we can square the equation of the parabola, giving x4 = cy 2 , or y 2 = (1/c)x4 , and then substitute this value for y 2 into the equation of the circle. We get !

1 d x − x = x4 . c c If we divide each side by x and multiply each side by c, we have !

d c − x = x3 c

d − cx = x3

x3 + cx = d,

so the original equation is in fact satisfied by the x-coordinate of the intersection point. Al-Khayy¯am¯ı noted here, without any indication of a proof, that this class of equations always has a single solution. In other words, the parabola and circle always intersect in one point other than the origin. The origin, though, does not provide a solution to the problem. Al-Khayy¯am¯ı’s remark reflects the modern statement that the equation x3 +cx = d always has exactly one positive solution. You should convince yourself of the truth of that statement by showing that the graph of y = x3 + cx − d always crosses the positive x-axis exactly once. Al-Khayy¯am¯ı treated each of his fourteen cases in the same manner. In those in which a positive solution does not always exist he gave a geometric condition for the existence. Namely, there are zero, one, or two solutions depending on whether the conic sections involved do not intersect or intersect at one or two points. His one failure in this analysis was in the case of the equation x3 + cx = bx2 + d, where he did not discover the possibility of three solutions.

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Exercises: 3 1. Show algebraically that one can solve the cubic equation √ x +d = cx by intersecting the 2 2 2 hyperbola y − x + (d/c)x and the parabola x = cy. Sketch the two conics. Find sets of values for c and d for which these conics do not intersect, intersect once, and intersect twice. In each case, draw the graph of the curve y = x3 −cx+d and show that the number of intersections of this curve with the positive x-axis are in agreement with the number of intersections of the conics. You may want to use a graphing calculator to do this.

2. Show algebraically that one can solve the cubic equation x3 + d = bx2 by√intersectthat if 3 d ≥ b, ing the hyperbola xy = d and the parabola y 2 + dx = db. Show √ 3 then the two conics do not intersect. Then, give examples when d < b of cases where the curves intersect twice, intersect once, or do not intersect at all. Determine conditions on the coefficients b and d which tell you whether there will be two, one, or no intersections. Consider in each case the graph of y = x3 − bx2 + d and its possible intersections with the positive x-axis. You can use a graphing calculator to assist you here. 3. Find two conic sections whose intersection will determine the solution(s) to the cubic equation x3 +cx = bx2 +d. Determine the conditions on the coefficients of this equation which will assure that it has three positive solutions. How do these conditions relate to the curves you have determined? You may want to use your graphing calculator to graph the conic sections as well as graph the curve y = x3 − bx2 + cx − d. You can then consider either the intersections of the conics or the intersections of y = x3 −bx2 +cx−d with the positive x-axis. 4. Find a pair of conic sections such that the x-coordinate of the intersection will give you the solution to the cubic equation x3 = bx2 + d. (Note that because the coefficients here are always positive, this cubic is different from the one considered in exercise 2.)

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Solving Cubic Equations Algebraically Teacher Notes Level: This activity is designed for use in the second year algebra course in high school. Materials: Historical background; student pages Time Frame: To cover this material thoroughly in class, a week would be necessary. However, students may work through the material at their own pace as an enrichment activity. Objectives: To extend studentsâ&#x20AC;&#x2122; knowledge of solving equations by considering the algebraic solution of cubic equations. To begin to develop the relationship between the solutions and the coefficients of a cubic polynomial equation. When to Use: This activity can be used after the students have a thorough understanding of the solution of quadratic equations, including the relationship between the solutions and the original coefficients. How to Use: This material can be covered in class by the teacher. In this case, the student pages will serve essentially as textbook material to be read before and after the class presentation. But the material could also be assigned for independent work or small group work. In either case, the historical background should be assigned as independent reading, and exercise set 2 should be done by the students before being discussed in class. Also, after the students have had time to think about the relationship between the roots and coefficients of a cubic, both when there is not a square term and when there is, their conjectures could be presented in class. Then a whole class discussion could help in working out the entire relationship.

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Solving Cubic Equations Algebraically Student Pages Historical Background Islamic mathematicians around the end of the first millennium noticed that Archimedes had solved a few special cubic equations geometrically by finding the intersection of two special curves. It was then the Persian mathematician and poet Omar Khayyam (1048–1131) who generalized this into a method by which he could solve any cubic equation. In his major work on algebra, Omar classified cubic equations and showed how in each case to find a pair of curves, either parabolas, hyperbolas, circles, or ellipses, such that the x-coordinate of their intersection would be a solution to the cubic equation. However, Omar was not entirely happy with this type of solution. As he wrote, “When, however, the object of the problem is an absolute number, neither we, nor any of those who are concerned with algebra, have been able to solve this equation — perhaps others who follow up will be able to fill the gap.” Omar was right — others did fill the gap. And so we come to sixteenth century Italy at the height of the Renaissance. In 1494, Luca Pacioli noted in his massive compendium of known mathematical results, the Summa de Arithmetica, Geometrica, Proportioni et Proportionalita, that there was not yet an algebraic solution to cubic equations in general. His implicit challenge to Italian mathematicians was picked up by several of them. Finally, sometime between 1500 and 1515, Scipione del Ferro (1465–1526), a professor at the University of Bologna, discovered an algebraic method of solving any cubic equation of the form x3 + cx = d. For example, he could show how to solve x3 + 6x = 20. We should note here that even in the Renaissance, people were uncomfortable with negative numbers. So just as they classified quadratic equations into three basic types, they also classified cubic equations into 13 different types, depending on which terms were on which sides of the equation. For example, they would not write an equation such as x3 − 6x = 20, but would instead write this as x3 = 6x + 20, and its solution method would likely be different from the method used to solve x3 + 6x = 20. So when del Ferro made his discovery, it was only a limited one. He only learned how to solve cubic equations of one specific type. This is important to the rest of the story. In modern academia, professors announce and publish new results as quickly as possible to ensure priority, so it may be surprising to learn that del Ferro did not publish, nor even publicly announce, his major breakthrough. But academic life in sixteenth-century Italy was far different from that of today. There was no tenure. University appointments were mostly temporary and subject to periodic renewal by the university senate. One of the ways a professor convinced the senate that he was worthy of continuing in his position was by winning public challenges. Two contenders for a given position would present each other with a list of problems, and, in a public forum some time later, each would present his solutions to the other’s problems. Often, considerable amounts of money, aside from the university positions themselves, were dependent on the outcome of such a challenge. As a result, if a professor discovered a new method for solving certain problems, it was to c 2004, M.A.A.

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his advantage to keep it secret. He could then pose these problems to his rivals secure in the knowledge that he would win. So although he did not publish his solution, del Ferro did disclose the solution to his pupil, Antonio Maria Fiore and to his successor at Bologna, Annibale della Nave. Although neither of these men published the solution either, word began to circulate in Italy that this old problem of solving cubic equations had been, or soon would be, solved. In fact, another mathematician, Niccolo Tartaglia of Brescia (1506–1559), boasted that he had discovered the solution to one type of cubic equation, the type x3 + bx2 = d. An example of this type would be the equation x3 + 5x2 = 28. In 1535, Fiore challenged Tartaglia to a public contest, hoping to win on the strength of his knowledge of the case x3 + cx = d. Each of the thirty problems he submitted for the challenge dealt with that class of cubic equations. For example, one of the problems read, “A man sells a sapphire for 500 ducats, making a profit of the cube root of his capital. How much is this profit?” If we set the capital equal to x3 , then the equation for this problem is x3 + x = 500. Tartaglia, however, was much the better mathematician than Fiore. He worked long and hard on finding a method to solve this type of cubic and, as he later wrote, on the night of February 12, 1535, he discovered the solution. Since he could now solve all of Fiore’s problems, and since Fiore was unable to solve many of Tartaglia’s questions — which dealt with other areas of mathematics — Tartaglia was declared the winner. The prize in this contest was not a university position, but 30 banquets prepared by the loser for the winner and his friends. Word of the contest and the new solutions of the cubic soon reached Milan, where Gerolamo Cardano (1501–1576) was giving public lectures in mathematics, supported by a grant from the will of a wealthy scholar for the instruction of poor youths. Cardano wrote to Tartaglia, asking that Tartaglia show him how to solve cubic equations so he could include the method, with full credit, in the arithmetic text Cardano was then writing. Tartaglia initially refused, but after many pleas and a promise from Cardano to introduce him and his inventions in artillery to the Milanese rulers, he finally came to Milan in early 1539. After getting Cardano to swear an oath that he would never publish Tartaglia’s solution — because he planned to publish them himself at a later date — Tartaglia divulged the secrets of three different forms of the cubic equation to Cardano in the form of a poem: When the cube and the things together Are equal to some discrete number, Find two other numbers differing in this one. Then you will keep this as a habit That their product should always be equal Exactly to the cube of a third of the things. The remainder then as a general rule Of their cube roots subtracted Will be equal to your principal thing. In the second of these acts, c 2004, M.A.A.

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When the cube remains alone, You will observe these other agreements: You will at once divide the number into two parts So that the one times the other produces clearly The cube of a third of the things exactly. Then of these two parts, as a habitual rule, You will take the cube roots added together, And this sum will be your thought. The third of these calculations of ours Is solved with the second if you take good care, As in their nature they are almost matched. These things I found, and not with sluggish steps, In the year one thousand five hundred, four and thirty, With foundations strong and sturdy In the city girdled by the sea. We will figure out the meaning of this poem below. But first, we want to finish up the story. Cardano had promised Tartaglia that he would not publish Tartaglia’s solution, and he kept his promise with regard to the arithmetic text he was just finishing. In fact, he sent Tartaglia a copy of that book, hot off the press, to prove his good faith. But Cardano was fascinated by the problem, and began to work on it himself. Remember that although Tartaglia showed him in verse the solution to three types of cubics, there were still many other types to work out. And over the next several years, he and his student Lodovico Ferrari (1522–1565) were able to work out the solutions to all of the types of cubic equations. Cardano, of course, did not want to break his oath, but he did want to make his solutions available. Acting on rumors of the original discovery by del Ferro, he and Ferrari traveled to Bologna and called on della Nave. The latter graciously gave Cardano permission to look at del Ferro’s original papers, and Cardano was able to verify that del Ferro had discovered the solution first. Cardano then decided he could publish the material on cubics. After all, he would not be publishing Tartaglia’s solution at all, but the solution discovered some twenty years earlier by a man now dead. So in 1545 Cardano published the solution methods for cubic equations in his most important mathematical work, The Great Art or The Rules of Algebra. Tartaglia was furious when Cardano’s work appeared. He felt he had been cheated of the rewards of his labor, even though Cardano did mention that Tartaglia was one of the original discoverers of the method of solving cubic equations. But Tartaglia’s protests availed him nothing. In fact, in an attempt to recoup some of his prestige, he had another public contest, this time with Ferrari. Tartaglia lost that contest, however, and died a broken man. To this day, in fact, the formulas providing the solution to cubic equations are known as Cardano’s formulas.

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Solving Cubic Equations We are finally prepared to tackle the solution of cubic equations. We will begin with equations of the form x3 = cx + d, equations which Cardano called “the cube equal to the first power and number.” In his book, Cardano states the rule as follows: When the cube of one-third the coefficient of x is not greater than the square of one-half the constant of the equation, subtract the former from the latter and add the square root of the remainder to one-half the constant of the equation and, again, subtract it from the same half. The sum of the cube roots of these two quantities constitutes the value of x. On the other hand, Tartaglia’s poem, in its second verse, tells us to divide the number into two parts, whose product is the cube of a third of the “things”, that is, the cube of one-third of the coefficient of x. The sum of the cube roots of these two parts is then the solution to the problem. Note that both Cardano and Tartaglia tell us to add the cube roots of certain numbers together. In order to show that the poem does imply Cardano’s rule, we need to show that these numbers are in fact the same for both men. We will do an example first. Let us try the equation x3 = 6x + 6. According to Cardano, we first must check that the cube of one-third the coefficient of x is not greater than the square of one-half the constant of the equation. (We will see the reason for this precaution shortly.) In this case, the coefficient of x is 6, so one-third of it is 2 and the cube of that number is 8. Since the constant of the equation is 6, one-half of it is 3 and the square of that is 9. Since 8 is less than 9, we can proceed with the solution. We now subtract the former, namely 8, from the latter, namely 9. We get 1. We take the square root of that, which is also 1, and add that to one-half the constant, 3. That gives us 4. We also subtract the 1 from one-half the constant, 3, and √ get√2. We then take the sum of the cube roots of these two quantities. Namely, x = 3 4 + 3 2. That is the solution to our equation. You might want to check this answer on your calculator. To do √ √ multiply it by 6 and add 6. You must that, you will have to calculate 3 4+ 3 2 in decimals, √ √ 3 3 then show that this answer is equal to the cube of 4 + 2. Now let us look at Tartaglia’s poem in this situation. He tells us that we want to divide the number, namely 6, into two parts, whose product is equal to the cube of one-third of the coefficient of x. The coefficient of x is 6, so one-third of that is 2, and its cube is 8. In other words, we need to divide 6 into two parts whose product is 8. The two parts of 6 which √ √satisfy this criterion are 4 and 2. Thus, according to Tartaglia, the solution is x = 3 4 + 3 2, the same solution we have already found. In this particular case, it was easy to follow Tartaglia’s rule. We can find two numbers whose sum is 6 and whose product is 8. But it is not always so easy. What Cardano’s procedure does is tell us explicitly how to find these numbers. So we see that Cardano actually had to do some significant work to turn Tartaglia’s poem into a usable rule. He certainly did not just copy his material from Tartaglia. To make it easier to use Cardano’s rule, we now want to turn it into a formula. So we begin with the algebraic equation x3 = cx + d. The rule says subtract the cube of onethird the coefficient of x from the square of one-half the constant. In algebraic terms, we must subtract ( 3c )3 from ( d2 )2 . We then take the square root of this difference and both c 2004, M.A.A.

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add it to and subtract it from one-half of the constant. We thus have two numbers: v u

d u d +t 2 2

c − 3

v u

d u d −t 2 2

and

−

c 3

Finally, our solution x is the sum of the cube roots of these two quantities. That is, Cardano’s formula for solving the cubic equation x3 = cx + d is x=

v u u u 3 d t

v u u t

d 2

c − 3

v u u u 3 d t

−

v u u t

d 2

−

c 3

Recall that Cardano also stated that the rule only applied when the cube of one-third of the coefficient of x is not greater than the square of one-half the constant. In other words, Cardano wants to insure that ( d2 )2 − ( 3c )3 is not less than 0. For if that quantity were negative, we could not take the square root of it, as the formula tells us we must. We cannot take the square root of a negative number. However, it turns out, and we will look at this later, that the formula still works even if the quantity inside the square root sign is negative. But in that case, we need to figure out how to deal with square roots of negative numbers. Cardano, in his book, does not just give the rule for determining the solution. He shows us why it works. So we too will derive his formula to show why it gives the solution to the cubic equation x3 = cx + d. Recall that Cardano’s rule, as well as Tartaglia’s poem, tells us that the solution x is the sum of two quantities. So let us write x = u + v. If we substitute in the equation x3 = cx + d, we get (u + v)3 = c(u + v) + d. We now rewrite the left side by using the binomial theorem for cubes and factoring. We get (u + v)3 = u3 + 3u2 v + 3uv 2 + v 3 = 3u2 v + 3uv 2 + u3 + v 3 = 3uv(u + v) + u3 + v 3 . If we now compare the right side of the expression just derived with the right side of the original equation, we get 3uv(u + v) + u3 + v 3 = cx + d. But since x = u + v, we can solve this equation if we can solve the two equations 3uv = c 3

u + v3 = d To solve this system of two equations in two unknowns, we recall that we know how to find a quadratic equation with two given solutions, assuming we know both the sum and the product of those solutions. In this case, the sum of u3 and v 3 is d. And since 3uv = c, we also have that uv = 3c and therefore that the product of u3 and v 3 is ( 3c )3 . Thus, the quadratic equation which has u3 and v 3 as solutions is y 2 − dy + ( 3c )3 = 0. We can solve this equation by the quadratic formula. We get the solution d± y= c 2004, M.A.A.

d2 − 4 2

3 c 3

. Funded by the N.S.F.

We can simplify this somewhat by breaking√apart the fraction and thinking of the 2 in the denominator of the right-hand term as 4. Then we use the rule that the quotient of square roots is the square root of the quotient. So we get d y= ± 2 or, if we note that

d2 4

d2 c − 4 3

is the square of d2 , v u

d u d y = ±t 2 2

−

c 3

We have therefore found that u3 and v 3 are given by the two values for y just found. It follows that

v u u u 3 d t

v u u t

d 2

c − 3

and v =

v u u u 3 d t

−

v u u t

d 2

−

c 3

Since the solution of the original cubic is the sum of u and v, we have now derived Cardano’s formula, as promised. The solution of x3 = cx + d is given by

v u u u 3 d t

v u u t

d 2

c − 3

v u u u 3 d t

−

v u u t

d 2

−

c 3

Let us try a few examples to illustrate the use of Cardano’s formula. We begin with = 6x + 9. Comparing this with the original equation x3 = cx + d, we have c = 6 and d = 9. Our formula gives x3

v u u 3 9 t

2 2 2

v u u 3 9 t

+ + +

s 2

9 2

s s s

6 − 3

81 − 23 + 4 81 32 − + 4 4

v u u 3 9 t

−

s 2

9 2

−

6 3

81 − 23 4

81 32 − 4 4

49 2 2 4 s s √ 7 7 √ 3 3 3 9 3 9 + + − = 8+ 1 = 2 2 2 2 = 2 + 1 = 3.

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49 + 4

v u u 3 9 t

−

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After all that calculation, we find that our answer is 3. We probably could have figured that out just by looking hard at the original equation. In any case, it is easy to check that 3 is correct, because 33 = 6 · 3 + 9. We now try an example where the formula does not give us quite so simple an answer, the equation x3 = 6x + 40. In this case, we have c = 6 and d = 40. Thus our formula gives the solution as x=

v u u 3 40 t r 3

40 2

6 − 3

v u u 3 40 t

−

40 2

−

6 3

202 − 23 + 20 − 202 − 23 q q √ √ 3 3 = 20 + 400 − 8 + 20 − 400 − 8 q q √ √ 3 3 = 20 + 392 + 20 − 392. =

20 +

We now have a solution to our cubic equation. The answer is a bit messy, but it is still an answer. We can determine the answer in decimals by using the calculator. We get q q √ √ 3 3 x = 20 + 392 + 20 − 392 √ √ = 3 20 + 19.79899 + 3 20 − 19.79899 √ √ 3 3 = 39.79899 + 0.20101 = 3.41421 + 0.58579 = 4. That is a bit strange. Our answer seems to be 4. Well, we can check that easily. 43 = 64 and 6 · 4 + 40 = 24 + 40 = 64, so, yes, 4 is a solution to our cubic equation. It is somewhat curious that 4 is expressed as the sum of two cube roots which cannot apparently be simplified algebraically. But there it is. In any case, we have certainly checked that our answer is correct. Let us try one more example: x3 = 12x + 36. Here, c = 12 and d = 36. The formula gives v v x=

u u 3 36 t r 3

36 2

12 − 3

r 3

u u 3 36 t

−

36 2

−

12 3

182 − 43 √ √ 3 3 = 18 + 324 − 64 + 18 − 324 − 64 q q √ √ 3 3 = 18 + 260 + 18 − 260. In this case, the solution is not an integer. We can calculate it using our calculator as follows: q q √ √ 3 3 x = 18 + 260 + 18 − 260 √ √ = 3 18 + 16.12452 + 3 18 − 16.12452 √ √ 3 3 = 34.12452 + 1.87548 = 3.24356 + 1.23321 = 4.47677. You can use your calculator to check that this result satisfies the original equation. =

18 +

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182 − 43 +

18 −

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Exercise Set 1: Solve each of the cubic equations 1 through 4 using Cardanoâ&#x20AC;&#x2122;s formula. 1. x3 = 9x + 12 2. x3 = 3x + 2 3. x3 = 9x + 28 4. x3 = 6x + 8 5. Recall that Cardano required for his formula to be correct that ( d2 )2 be at least as large as ( 3c )3 , because otherwise, we get a negative number inside the square root sign in the formula. But what happens if this condition is not satisfied. Consider the equation x3 = 30x+36. First, find a solution to this equation by inspection. The solution is a relatively small whole number. Now, apply the formula. What do you get? Can you convince yourself that the answer you get from the formula is the same as the answer you found by inspection?

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As we have seen, Cardano stated that his method only works if the cube of one-third the coefficient of x is not greater than the square of one-half the constant of the equation. Let us try an example to see what happens when this condition does not hold. We try to solve x3 = 15x + 4. Here the cube of one-third the coefficient of x is 53 = 125, while the square of one-half the constant is 22 = 4. And 125 is greater than 4. So what does Cardano’s formula give in this case. We set c = 15 and d = 4. We get

v u u 3 4 t r 3

s 2

4 2

15 − 3 r 3

v u u 3 4 t

−

s 2

4 2

−

15 3

22 − 53 + 2 − 22 − 53 q q √ √ 3 3 = 2 + 4 − 125 + 2 − 4 − 125 q q √ √ 3 3 = 2 + −121 + 2 − −121. =

Now we see the problem. The answer is expressed in terms of the square root of −121, and we know that negative numbers do not have real square roots. On the other hand, we can easily find the solution to x3 = 15x + 4 by inspection. If we try a few small whole numbers, we see that x = 4 is certainly a solution. After all, 43 = 64, and 15 · 4 + 4 = 64 as well. q q √ √ So the question is how does 3 2 + −121+ 3 2 − −121 equal 4. If you plug the value given by the formula into most calculators, it will give an error message. So this problem is somewhat different than the one we looked at earlier, where 4 was also expressed as the sum of cube roots, but of cube roots of real numbers we could check on our calculator. What did Cardano do when he ran into negative numbers inside square root signs? He looked for other ways of solving the cubic equation. In fact, we will look at a trigonometric method in another section of this module. But we will also look at what we can do to show why this strange expression coming from our formula does in fact equal 4. In doing so, we will have to introduce an entirely new kind of number. But before we do this, we want to look at how we can solve other cubic equations, equations which are not in the form cube equal to first power and number. For Cardano and his contemporaries, negative numbers were still a problem. Negative numbers could certainly not be coefficients in cubic equations, although Cardano was willing to admit negative numbers as solutions, even though he called them “false” solutions. So Cardano was forced to consider three different kinds of cubics in the situation where there was no square term (and many more when there was a square term). That is, besides the case “cube equal to first power and number” that we have already considered, Cardano had separate treatments of the cases “cube and first power equal to number” (x3 + cx = d) and “cube and number equal to first power” (x3 + d = cx). Exercise Set 2: We will present here Cardano’s rules for solving these cases and compare them with the first and third verses of Tartaglia’s poem. Your task is to develop formulas for the two cases “cube and first power equal to number” and “cube and number equal to first power.” c 2004, M.A.A.

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So first, reread Tartaglia’s first verse. In the case Tartaglia is considering there, the case x3 + cx = d, Cardano states the rule as follows: Cube one-third the coefficient of x; add to it the square of one-half the constant of the equation; and take the square root of the whole. You will add to this one-half the number you have already squared and also subtract from it one-half the same. Then subtracting the cube root of the latter quantity from the cube root of the former, the remainder is the value of x. 1. Use Cardano’s rule to solve x3 + 6x = 20. Show how his rule follows Tartaglia’s poem. 2. Turn Cardano’s rule into a formula similar to Cardano’s formula above for the earlier case. 3. Use your formula to solve x3 + 3x = 10 and also x3 + 6x = 2. 4. Derive Cardano’s formula for the case x3 + cx = d in a manner analogous to our derivation of Cardano’s formula for the case x3 = cx + d. But instead of setting x = u + v, set x = u − v. Now reread Tartaglia’s third verse. Note that Tartaglia did not really give a method of solution for this third case, “cube and number equal to first power.” He just suggested that you could solve it if you studied the second case, the case “cube equal to first power and number.” So Cardano had to work to discover the solution here. Here is Cardano’s rule to solve x3 + d = cx. First, find the solution to the corresponding equation y 3 = cy + d. Next take three times the square of one-half of this solution and subtract it from the coefficient of the x. Then the square root of the remainder added to or subtracted from one-half of the solution to y 3 = cy + d gives the solution for x. We follow Cardano’s example of this. To solve x3 + 3 = 8x, first we need to solve y 3 = 8y + 3. The solution to this, found by inspection, is y = 3. We then take one-half of this solution, 1 12 , square it to get 2 14 , and multiply this value by 3, giving 6 34 . We next subtract this from the coefficient of x, namely 8, to get 1 14 . We take the square root of this remainder and both add it to and subtract it from the value 1 12 already calculated. Thus, we get two solutions to our equation

+ 3 = 8x: x =

1 12

1 14

and x =

1 12

−

1 14 .

5. Turn Cardano’s rule into a formula for solving x3 + d = cx. 6. Use this formula to solve x3 + 2 = 5x. (First, discover a solution to y 3 = 5y + 2 by inspection.) 7. Use this formula to solve x3 + 4 = 15x. (First, find a solution y 3 = 15y + 4 by inspection.) 8. It is perhaps curious that Cardano does not mention the obvious solution to x3 + d = cx, given a solution y to y 3 = cx + d. This is the solution x = −y. Of course, since Cardano was unhappy using negative numbers, perhaps he wouldn’t mention this, even by calling it a “false” solution. Show in Cardano’s example as well as problems 6 and 7 that the negative of the answer you found to the equation “cube equals first power plus number” is a solution to “cube plus number equals first power.” c 2004, M.A.A.

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9. If you use the information from problem 8, along with the solutions you found in problem 6, you will have discovered three solutions to x3 + 2 = 5x. What are they? 10. Using the information from problems 8 and 9, determine three solutions to x3 = 5x +2. 11. Recall from your study of quadratic equations that in the equation x2 + bx + c = 0, the sum of the two roots was −b and the product was c. Write each of the two equations x3 + 2 = 5x and x3 = 5x + 2 in the form x3 + cx + d = 0. In each case, find the product of the three solutions and the sum of the three solutions. Compare these answers to the coefficients in your equation. Make a conjecture about the relationship of the roots to the coefficients in a cubic equation. 12. Find three solutions to each of x3 +3 = 8x, x3 = 8x+3, x3 +4 = 15x, and x3 = 15x+4. Check the conjecture you make in problem 11 in these four situations. Modify your conjecture if necessary. 13. Show that in the three equations of the form x3 = cx + d that we considered here (in Cardano’s example as well as in problems 6 and 7), if you try to solve these using Cardano’s formula, you get, in each case, a negative number inside the square root sign. On the other hand, recall that we solved x3 = 6x + 9 by using that formula. Show that if you try to use Cardano’s rule from problem 5 on x3 + 9 = 6x, you run into negative numbers inside the square root sign from that rule.

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As we saw in the exercises, Cardano used different methods to solve different types of cubic equations. But since we can deal with negative numbers, it is simplest for us to combine all of Cardano’s cases – and even a case he did not consider – into one equation, the equation x3 + cx + d = 0. Here we will allow b or c or both to be negative. And since our derivation of Cardano’s original solution formula 5.11 did not depend on the sign of c or d, we can simply rearrange the terms in our equation and use the same formula. We begin by changing the equation x3 + cx + d = 0 into Cardano’s form, x3 = −cx − d. We then apply the formula, replacing the c by −c and the d by −d. We get

v u u u 3 −d t

v u u t

−d 2

−c − 3

v u u u 3 −d t

−

v u u t

−d 2

−

d But since the square of −d 2 is equal to the square of 2 and the cube of negative of the cube of 3c , we can rewrite this solution as

v u u u d 3 t −

v u u t

d 2

c + 3

v u u u d 3 t −

−

v u u t

d 2

−c 3

is equal to the

c 3

Since the cube root of a negative number is the negative of the cube root of the number, and since it “looks better” if we put the d2 term after the square root term, we will finally rewrite this in the form

vv uu uu u 3 t t

d 2

c + 3

d − − 2

vv uu uu u 3 t t

d 2

c 3

d + . 2

We now have a formula that will allow us to solve any cubic equation which does not have a square term. (We will see what to do about square terms shortly.) We only have to make sure that the number under the square root sign is not negative. Examples: 1. We solve x3 + 6x + 20 = 0 by the new formula. Here c = 6 and d = 20. We get

vs u u 3 20 2 t rq 3

6 + 3

20 − − 2 rq

vs u u 3 20 2 t

6 3

20 2

102 + 23 − 10 − 102 + 23 + 10 q√ q√ 3 3 = 108 − 10 − 108 + 10. =

If we calculate this value on our calculator, we get √ √ 3 3 x = 0.39230 − 20.39230 = 0.73205 − 2.73205 = −2. c 2004, M.A.A.

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We easily check that −2 is a solution of the equation: (−2)3 + 6(−2) + 20 = −8 − 12 + 20 = 0. 2. We solve x3 + 6 = 6x. To use the new formula, we first must rewrite this in the form x3 − 6x + 6 = 0. Then, c = −6 and d = 6. We get x=

vs u u 3 6 2 t rq

−6 + 3

6 − − 2

vs u u 3 6 2 t

−6 + 3

6 2

32 − 23 − 3 − 32 − 23 + 3 q√ q√ √ √ 3 3 = 1−3− 1+3= 31−3− 31+3 √ √ √ √ 3 3 3 = 3 −2 − 4 = − 2 − 4.

Although this is√a perfectly nice solution, if we choose, we can calculate this value on √ 3 3 our calculator: 2 + 4 = 1.25992 + 1.58740 = 2.84732. So our solution is x = −2.84732. We can check that this is a solution to the original equation. 3. We solve x3 + 3x = 10. Again, to use the formula, we must rewrite this in the form x3 + 3x − 10 = 0, so c = 3 and d = −10. We then get x=

vs u u 3 −10 2 t rq 3

3 3

−

−10 − 2

vs u u 3 −10 2 t

3 3

−10 2

(−5)2 + 13 + 5 − (−5)2 + 13 − 5 q√ q√ 3 3 = 26 + 5 − 26 − 5. =

Again, although we could leave the answer in radical form, we can also write this in decimals: √ √ x = 3 5.09902 + 5 − 3 5.09902 − 5 √ √ 3 3 = 10.09902 − 0.09902 = 2.16152 − 0.46264 = 1.69888.

Exercise Set 3: Solve each of the following cubic equations by using the new formula. 1. x3 = 6x + 20 2. x3 + 6x = 16 3. x3 + 64 = 18x 4. x3 + 24 = 9x 5. x3 + 6x + 2 = 0 c 2004, M.A.A.

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Although we have looked at several types of cubic equations, we have not considered equations which have square terms. Cardano, of course, does consider these, but gives a different procedure for each one. For example, for the equation “cube equal to square and number” (in our notation, x3 = bx2 + d), Cardano gives the following rule: Add the cube of one-third the coefficient of x2 to one-half the constant of the equation and square the sum of these. From this square subtract the cube of the square of one-third the coefficient of x2 and add or subtract the square root of the remainder to or from the sum which was squared. Then add the cube roots of these two quantities together and add to their sum one-third the coefficient of x2 . The result is the value of x. Cardano’s rules in other situations are even more complicated, so we will not mention them. But we can subsume all of his rules under one procedure. The idea is to eliminate the x2 term by a simple substitution and therefore reduce the equation to one which we already know how to solve. We illustrate this by the example x3 = 6x2 + 20. We let x = y + 2, where the 2 is the one-third of the coefficient of x2 mentioned in Cardano’s rule. Thus, we replace the x in both places in the equation by y + 2. The substitution gives us (y + 2)3 = 6(y + 2)2 + 20. If we expand the left side by the binomial theorem for cubes, we get y 3 + 6y 2 + 12y + 8. Similarly, the right side becomes 6(y 2 + 4y + 4) + 20 = 6y 2 + 24y + 44. So our equation in x becomes the following equation in y: y 3 + 6y 2 + 12y + 8 = 6y 2 + 24y + 44 Subtracting now 6y 2 from each side, we get an equation without any square term, namely y 3 + 12y + 8 = 24y + 44, which reduces to y 3 = 12y + 36. Recall that we already solved this equation. The result is y=

q 3

q √ √ 3 18 + 260 + 18 − 260.

To get our valueqfor x, we recallqthat x = y + 2. Thus the solution to our original equation √ √ 3 3 is given by x = 18 + 260 + 18 − 260 + 2. This method of replacing x by y plus one-third the coefficient of x2 (or sometimes y minus one-third the coefficient of x2 – depending on which side of the equation the square term is), will always eliminate the square term. Thus it will always give us an equation to which we can apply our previous formulas. Given the solution for y from one of those formulas, we merely have to add to it or subtract from it one-third the coefficient of x2 to give us the solution x of the original equation. So rather than go through all of Cardano’s rules, we will simply give you a few exercises on which to practice this procedure.

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Exercise Set 4: Solve each of the following cubic equations. In some of these, the you will already have solved the equation without the square term to which the given equation reduces by using the appropriate substitution. 1. x3 + 6x = 6x2 + 5 2. x3 = 3x2 + 3x + 35 3. x3 + 18x = 9x2 + 12 4. x3 + 51x = 12x2 + 76 5. x3 + 3x2 = 5x + 4 6. x3 + 6x2 = 7 7. In each of problems 5 and 6, you should be able to determine three solutions to the cubic equation. In each case, rewrite the equation in the form x3 + bx2 + cx + d = 0, then determine the sum and the product of the three solutions. Compare these answers to the coefficients. What do you find? Does your earlier conjecture about the three solutions and the coefficients apply in this case? If not, can you modify your conjecture so that it applies in both situations? 8. In each of problems 1 through 4, you probably only were able to find one solution. Yet, somehow we feel that a cubic equation should have three solutions. Does the conjecture you made in problem 7 enable you to determine the other solutions? What do you know about these solutions? 9. Turn Cardanoâ&#x20AC;&#x2122;s rule for solving equations of the form x3 = bx2 + d into a formula. Then turn the method used in the example x3 = 6x2 + 20 into a formula. Check that the two formulas are equivalent.

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Answers to Exercises Exercise Set 1

Exercise Set 3 1.

q√

2. 2

q√

72 + 8 −

3. 4

q√

808 − 32 −

q√

117 − 12 −

q√

√ √ 3 2− 34

√ √ 3 9+ 33

q 3

q √ √ 3 8+ 4− 8

Exercise Set 2 1.

q√ 3

2. x = 3.

108 + 10 −

q√

+ (c/3)3

(d/2)2

y 2

q√ 3

92 + 10 −

q√ 3

92 − 10

72 − 8 808 + 32 117 + 12

1089 − 10

√ √ 3 4− 32

5. x =

+ d/2 −

rq 3

(d/2)2 + (c/3)3 − d/2 Exercise Set 4

c − 3( y2 )2

1. 5

√ 2

6. −1 ±

2. 5

√ 7. 2 ±

√ √ 3 9+ 33+3

√ 9. 2, −1 ±

4. 4

5. −4,

√ 10. −2, 1 ± 12a. −3, 1 12 ± 12b. 3, −1 12 ±

1 14

1 2

6. 1, − 72 ±

1 34

1 2

√ 21

1 14

√ 12c. −4, 2 ±

3 √

12d. 4, −2 ±

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Introduction to Complex Numbers Teacher Notes Level: This activity is designed for use in the second year algebra course in high school. Materials: Student pages Time Frame: One or two class periods, with some homework and discussion. Objective: To introduce the notion of a complex number in the context in which this concept was originally invented. When to Use: This activity can be used after the students have worked through the material on the solution of cubic equations via Cardano’s formula. How to Use: This material can be covered in class by the teacher. In this case, the student pages will serve essentially as textbook material to be read before and after the class presentation. The material could also be assigned for independent or small group work. Answers to Exercises: Exercise Set 1 1. 4i

Exercise Set 2 9. 1

12 13

5 13 i

2. 5 + 11i

10. 5 + 12i

4 3

− 23 i

3. 1 + 4i

11. i

13 10

4. 5 − 5i

12. −2 + 2i

4. −2i

5. −15

13. −1

7 + 5. − 13

6. 10 + 11i

14. 1

7. 58

15. i17 = i, i99 = −i

8. −i

16. ω 2 = − 12 −

−

1 10 i

24 15 i

Exercise Set 3

√ 3 2 i

1. x = −1 ± 2i 2. x =

1 2

√ 3 2 i

3. x = 2 ± i 4. x = − 32 ± c 2004, M.A.A.

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Introduction to Complex Numbers Student Pages Recall that we have seen examples of cases where the use of Cardano’s formula for solving cubic equations of the form x3 = cx + d led us to negative numbers inside of square root signs, even where we knew in advance what the solution must be. Generally, when we solve quadratic equations using the quadratic formula, we decide that there are no solutions if there is a negative number inside the square root sign of that formula. But in the case of the equation x3 = 15x + 4, we cannot say that. We know that the solution is x = 4, yet we also saw that Cardano’s formula expresses this value as x=

q 3

√

−121 +

q 3

√

2−

−121.

So our question becomes, how can we justify the use of square roots of negative numbers in this situation? Can we make some sense of them? We know that Cardano’s formula works – because it gave us correct solutions in the cases we have tried, at least where there were no square roots of negative numbers. So we would like to show why it works even under those somewhat unusual circumstances. Cardano himself really did not know what to do in this situation. He worked out ways to get around the dilemma in certain cases, but was certainly not entirely happy with his results. It remained for another Italian mathematician, Rafael Bombelli, to solve this problem, in his own algebra text published in 1572. In another section of this module, we will see an alternative solution using trigonometry. To understand Bombelli’s solution, it is best first to look at a somewhat different problem. Recall that in solving the equation x3 = 6x + 40, an equation for which we know that the solution is x = 4, we found that Cardano’s formula expressed the solution in the form x=

q 3

20 +

q √ √ 3 392 + 20 − 392.

Although we in fact used our calculators to show that this expression is equal to 4, let us try to find another way of showing √ this, a way which Bombelli himself worked out. We assume that the quantities 20 ± 392 are the cubes of other algebraic quantities. What kinds of quantities could they be cubes of? It seems natural to assume they are √ √ cubes of b√or a − b, quantities that look something like them, namely, quantities of the form a + √ for some a and b. So let us assume that 20 + 392 is equal to the cube of a + b and see where that leads us. √ √ √ √ First, we note that if 20 + 392 = (a + b)3 , then 20 − 392 = (a − b)3 . If we multiply these two equations together, we get (20 +

√ √ √ √ 392)(20 − 392) = (a + b)3 (a − b)3 ,

or 400 − 392 = (a2 − b)3, c 2004, M.A.A.

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or, 8 = (a2 − b)3 , or, finally, a2 − b = 2. √

√ Next, we expand the right side of the equation 20 + binomial formula for cubes. We get √ 20 +

392 = (a +

b)3 , using the

√ √ 392 = a3 + 3a2 b + 3ab + b b.

If we then equate the parts of this equation with no square roots, we get a3 + 3ab = 20. We now have two equations in the two unknowns a and b. These equations are simple enough for us to solve by inspection. In fact, the first equation shows us that a2 > 2, while the second equation tells us that a3 < 20. The only integer a which satisfies both of these conditions is a = 2. If we substitute that value into either equation, we find that b = 2. We check easily that the values a = 2, b = 2 in fact satisfy both equations. √ 3 √ √ 392 = (2 + 2) and also that 20 − 392 = We have therefore shown that 20 + q q √ √ √ √ 3 √ 3 3 (2 − 2) . It follows that 20 + 392 = 2 + 2 and that q20 − 392 =q2 − 2. √ √ 3 3 Therefore, the solution to our original equation, namely, x = 20 + 392 + 20 − 392, √ √ can be written in the form x = (2 + 2) + (2 − 2) = 4. We have thus shown that the solution we knew was correct, namely x = 4, does come out of Cardano’s formula by some kind of algebraic manipulation. Bombelli played with many examples of solutions of this type. He generally used trial and error to find the values for a and b. Sometimes this worked, and other times it did not. There is no general way to simplify answers given as the sum of two cube roots of binomial expressions. Nevertheless, the partial success encouraged Bombelli to try the same thing when the value inside the square root sign was negative. q q √ √ 3 We return to the solution x = 2 + −121 + 3 2 − −121 of the equation x3 = 15x + 4, an equation whose solution we know √ to be x = 4. As in the example just discussed, we will assume, with Bombelli, that 2 + −121 is a cube. But the cube of what? It would seem that it would cube of an expression of the same type. So Bombelli as√ have to be a √ as before. sumed that 2 + −121 = (a + −b)3 and went through the same calculations √ The only problem is that during √ these calculations we have to multiply −121 by itself. So, without discussing what −121 “really” is, we will just assume that if we square it, we will get −121. In other words, we will assume that the normal algebraic rules apply to these mysterious square roots of negative numbers. √ √ √ 3 , then 2 − −121 = −121 = (a + −b) As before, then, if we assume that 2 + √ (a − −b)3 . If we multiply these two equations together, we get √ (2 + c 2004, M.A.A.

−121)(2 −

√ √ √ −121) = (a + −b)3 (a − −b)3 , 81

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or 4 − (−121) = (a2 − (−b))3 , or 125 = (a2 + b)3 , or, finally, a2 + b = 5. √ √ Next, as before, we expand the right side of the equation 2 + −121 = (a + −b)3 , using the binomial formula for cubes. We get √ √ √ √ √ 2 + −121 = a3 + 3a2 −b + 3a(−b) + (−b) −b = a3 + 3a2 −b − 3ab − b −b. If we then equate the parts of this equation with no square roots, we get a3 − 3ab = 2. Again we have two equations to solve in two unknowns. And again, we solve them by inspection. The first equation implies that a2 < 5, while the second shows that a3 > 2. The only possible value for a is then a = 2. If we substitute this value into either of the other equations, we find that b = 1. It follows that the values a√= 2, b = 1 satisfy √ both equations, and√we may conclude, with Bombelli, that 2 + −121 = (2 + −1)3 and √ along also that 2 − −121 = (2 − −1)3 . We can therefore rewrite the solution to the original equation x3 = 15x + 4. We get x=

q 3

q √ √ √ √ 3 2 + −121 + 2 − −121 = (2 + −1) + (2 − −1) = 4.

Since we already knew that the solution was x = 4, we may now be prepared to accept the validity of dealing with these square roots of negative numbers. For Bombelli, this computation, and others like it, convinced him that there was a reason to use such numbers. But Bombelli did not given any philosophical discussion as to the meaning of these numbers. In fact, he wrote, “The whole matter seemed to rest on sophistry rather than on truth.” Despite these doubts, Bombelli nevertheless went on to elaborate on the arithmetic of expressions involving square roots of negative numbers, expressions that today are known as complex numbers. √ Following the spirit, if not the exact notation, of Bombelli, we will now designate −1 by the letter i. That is, we will create a new “number,” the square root of −1, a number which has the normal properties of square roots, that when you multiply it √ by itself, √ you √ get the original number back. √ So we will assume the property of −1 that −1 −1 = −1, or, using i to represent −1, that i2 = −1. The only other property of i we will assume is that it obeys standard algebraic rules. So, for example, 3i + 7i = 10i, q i + i = 2i, √ √ √ √ and 2i − 5i = −3i. Also, since −1 = i, we have −4 = 4 · (−1) = 4 · −1 = 2i. √ √ √ √ √ √ √ Similarly, −3 = 3i, −12 = 12i = 2 3i, and, in general, −b = bi. We will further assume that we can “add” any multiple bi of i to an ordinary real number a to form what is called a complex number, a “number” of the form a + bi. The c 2004, M.A.A.

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plus sign here does not have the same meaning as our ordinary plus sign which tells us to add two real numbers, because here we cannot combine the two quantities a and bi to get a similar type of quantity. So the plus sign is only a symbol that connects the real number a to the imaginary number bi to form the complex number a + bi. In any case, we now have a new set of “numbers”, the complex numbers, to which we can apply the four basic operations of arithmetic. To do this, we will assume with Bombelli, that we can use the same algebraic rules on complex numbers that we use with real numbers and ordinary algebraic expressions. We begin with addition and subtraction. To add or subtract two complex numbers, we add or subtract the real and imaginary parts separately. (Note that a is called the real part and b is called the imaginary part of the complex number a + bi.) Thus, (3 + 4i) + (5 + 7i) = (3 + 5) + (4i + 7i) = 8 + 11i. Similarly, (2 − 6i) + (−4 + 3i) = −2 − 3i and (5 − 4i) − (2 + 2i) = 3 − 6i. The cardinal rule for multiplication is that i2 = −1. We then get (3i) · (2i) = 6i2 = 6 · (−1) = −6 and (−2i) · (4i) = −8i2 = (−8)(−1) = 8. For multiplying arbitrary complex numbers together, we use the distributive law. So (2 + 3i) · (5 + 7i) = 10 + 14i + 15i + 21i2 = 10 + 29i − 21 = −11 + 29i. Also, (3 − 4i) · (5 + 2i) = 15 + 6i − 20i − 8i2 = 15 − 14i + 8 = 23 − 14i. There is no particular reason to stick to integers as coefficients. We can multiply 2 1 3 − 2 i to get

1 2

+ 32 i by

3 3 13 3 1 1 1 3 1 3 2 1 + i · − i = − i + 1i − i2 = + i + = + i. 2 2 3 2 3 4 4 3 4 4 12 4

We could even have irrational numbers: √ √ √ √ √ √ (4 + 2 i) · (3 − 2 2 i) = 12 − 8 2 i + 3 2 i − 4i2 = 12 − 5 2 i + 4 = 16 − 5 2 i. As a final example of multiplication, we multiply 3 + 2i by 3 − 2i. We get (3 + 2i) · (3 − 2i) = 9 − 6i + 6i − 4i2 = 9 + 4 = 13. In this case, the imaginary terms canceled each other out and our result was a real number. A moment’s thought should convince you that this will always happen when we multiply a + bi by a − bi, whatever a and b are. The result is (a + bi) · (a − bi) = a2 − abi + abi − b2 i2 = a2 + b2 . Because the product of a + bi by a − bi is always a real number — and, in fact, a nonnegative real number — we give a − bi a special name. The complex number a − bi is called the complex conjugate of a + bi. You should note that not only is the product of a complex number by its complex conjugate (or just “conjugate” for short) a real number, but also the sum of a number and its complex conjugate is real. That is, (a+bi)+(a−bi) = 2a. c 2004, M.A.A.

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That the sum of a complex number and its conjugate is real is what we used in the example above on solving cubic equations using Cardano’s formula. The Cardano formula gave the solution to x3 = 15x + 4 as the sum of the cube roots of a complex number and its conjugate. We calculated that the cube root of the conjugate was in fact the conjugate of the cube root, a result that is true in general. So when Cardano’s formula gives us complex numbers inside the cube root sign, the real solution to the equation is given as the sum of a complex number and its conjugate. Exercise Set 1: Perform the indicated operations on the given complex numbers: √ 1. −16 2. (3 + 4i) + (2 + 7i) 3. (3 + 5i) − (2 + i) 4. (2 − 7i) − (−3 − 2i) 5. (3i) · (5i) 6. (4 + i) · (3 + 2i) 7. (3 − 7i) · (3 + 7i) 8. i3 (This simply means to multiply three copies of i together.) 9. i4 10. (3 + 2i)2 11. (

√ 2 2

√ 2 2 2 i)

12. (1 + i)3 √ 3 3 2 i) √ (− 12 + 23 i)3

13. ( 12 + 14.

15. You have calculated i2 , i3 , and i4 . What is i5 ? What happens with higher powers of i? Calculate i17 and i99 without actually doing any multiplication. √

16. Note that problem 14 shows that the cube of ω = − 12 + 23 i is equal to 1. In other words, ω is a cube root of 1. We know that 1 is also a cube root of 1. So we have two solutions of the cubic equation x3 = 1. Find a third solution, which is also a complex number. You may want to use the rules you discovered about the sums and/or products of the roots of a cubic equation. c 2004, M.A.A.

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We can now consider division of complex numbers. To divide 4i by 2 is straightforward. We just divide the 4 by 2 to get the answer 2i. Similarly, to divide 15 + 9i by 3, we just divide each term by 3 to get 5 + 3i. If the divisor does not divide the two terms evenly, our result will have fractions in it, but that is not surprising in a division problem. For example 3 − 4i divided by 5 is just 35 − 45 i. In the cases just considered, the divisor was a real number. What happens when the divisor has a non-zero imaginary part? For example, how do we divide 5 + 6i by 2 + i. To do this, we use the fact that the product of a complex number with its conjugate is real. So we set up the division formally as a quotient and multiply both the numerator and the denominator by the complex conjugate of 2 + i, namely 2 − i. We get 5 + 6i (5 + 6i)(2 − i) 10 − 5i + 12i − 6i2 16 7 16 + 7i = = = + i. = 2 2+i (2 + i)(2 − i) 4−i 5 5 5 For another example, let us divide 2 by 3 − 2i. As before, we multiply numerator and denominator in the quotient by the complex conjugate of the divisor, namely 3 + 2i. We have 2 2(3 + 2i) 6 + 4i 6 4 = = = + i. 3 − 2i (3 − 2i)(3 + 2i) 13 13 13

Exercise Set 2: Perform the divisions of complex numbers as indicated: 1. (2 + 3i) ÷ (3 + 2i) 2. (4 − 2i) ÷ 3 3. (5 − 2i) ÷ (3 + i) 4. 2 ÷ i 5. (3 + 4i) ÷ (3 − 4i)

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It is interesting to note that complex numbers were essentially invented by Bombelli to solve cubic equations even though the quadratic formula under certain circumstances gives us the square root of a negative number. For millennia, mathematicians had said in those cases that there was no solution. But in the case of the cubic equation, that answer was not possible. After all, we have seen examples where the solution was known to exist. It was just a matter of figuring out how to write the solution in terms of complex numbers. But Bombelli in his algebra text did realize that he could use complex numbers to solve the previously unsolvable cases of quadratic equations. For example, if we use the quadratic formula to solve the equation x2 − 8x + 20 = 0, we get −(−8) ± x=

(−8)2 − 4 · 20 2

8±

√ √ 8 ± −16 4i 64 − 80 = =4± = 4 ± 2i. 2 2 2

We thus have two solutions, 4 + 2i and 4 − 2i to the quadratic equation x2 − 8x + 20 = 0. Both of these solutions are complex numbers, and they are complex conjugates of one another. Furthermore, the solutions satisfy the standard rule relating them to the original coefficients. Namely, the sum of the two solutions is 8, the negative of the coefficient of x in the equation, while the product of the two solutions is (4 + 2i)(4 − 2i) = 16 + 4 = 20, the constant term of the equation. Let us try one further example. We solve x2 + x + 1 = 0. The quadratic formula gives us √ √ √ √ −2 ± 1 − 4 1 1 −3 3 −2 ± 12 − 4 · 1 = =− ± =− ± i. x= 2 2 2 2 2 2 √

√

Again, we have two complex solutions: x = − 12 + 23 i and x = − 12 − 23 i. And as before, the sum of the two solution is −1, the negative of the coefficient of x, while the product of the two solutions is 14 + 34 = 1, the constant of the equation. We note here that these two complex numbers have the property, which you may have discovered in an earlier exercise, that their cubes are equal to 1. Also, you should check that the second of these two numbers is the square of the first number. So, if we designate the first of these two solutions as ω, we now have three cube roots of 1, namely, 1, ω, and ω 2 . Exercise Set 3: Solve each of the quadratic equations in problems 1 through 4 to get two complex solutions: 1. x2 + 2x + 5 = 0 2. x2 − x + 1 = 0 3. x2 − 4x + 5 = 0 4. x2 + 3x + 6 = 0 5. In each of the equations above, check that the sum of the two solutions is the negative of the coefficient of x and that the product of the two solutions is the constant term. c 2004, M.A.A.

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We have seen that complex numbers help us to find the real solutions to cubic equations using Cardano’s formula. However, Cardano’s formula only seems to give one solution. We see in the next eight exercises that in fact the formula can be modified to give us three solutions. 6. We solve the equationqx3 = 21x + 20.qShow that Cardano’s formula gives us the solu√ √ tion in the form x = 3 10 + 9 −3 + 3 10 − 9 −3. 7. We √ try to simplify the answer √ in 6 by assuming that there √ is a complex number √ a+ b −3 whose cube is 10 + 9 −3. Then the cube of a − b −3 will be 10 − 9 −3. Using Bombelli’s method, show that this assumption leads to the following two equations in the two unknowns a and b: a2 + 3b2 = 7 and a3 − 9ab2 = 10. 8. It is easy enough to see that there are no positive integer solutions to the two equations from 7. However, there are negative and fractional solutions. So show that a = −2, b = 1 is a solution and also that a = 52 , b = 12 and a = − 12 , b = − 32 are solutions. √ √ √ 9. Using the results of 8, show that −2 + −3, 52 + 12 −3, and − 12 − 32 −3 are three √ distinct complex numbers whose cubes are equal to 10 + 9 −3. √ −3 are the complex 10. Using the results of 9 and the fact that the cube roots of 10 − 9 √ conjugates of the cube roots of 10 + 9 −3, find three distinct solutions to the equation x3 = 21x + 20 by substituting appropriately in the solution from 6. √ √ 11. Let us √rewrite −2 + −3 as α = −2 + 3 i. Show that the product of α with ω = √ √ other cube roots of 10 + 9 −3. Further, − 12 + 23 i is equal to − 12 − 32 3 i, one of the √ √ show that the product of α with ω 2 = − 12 − 23 i equals 52 + 12 3 i, the third cube root √ of 10 + 9 −3. 12. Show why, if you know one cube root α of a complex number, then there are two other cube roots, namely ωα and ω 2 α, where ω is defined in 11 and is, as we have seen earlier, a cube root of 1. 13. We know that one solution of the cubic equation x3 = 21x + 20 is x = −4, which we can write as q q √ √ √ √ 3 3 10 + 9 −3 + 10 − 9 −3 = (−2 + 3 i) + (−2 − 3 i). Now, show that if we multiply the first term in this sum by ω and the second by ω 2 , we get the solution x = −1. Finally, show that if we multiply the first term in the sum by ω 2 and the second by ω, we get a third solution x = 5. The previous eight exercises form one example of the following result. If Cardano’s for√ √ 3 3 mula gives us one solution of a cubic equation in√the form x = β + λ, then √ the two √ √ 3 3 3 2 2 other solutions of the cubic equation are x = ω β + ω λ and x = ω β + ω 3 λ. You can demonstrate the truth of this statement by using Cardano’s formula to write β and λ explicitly and then showing that the two values mentioned are in fact solutions to the original cubic. c 2004, M.A.A.

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Solving Cubic Equations Using Trigonometry Teacher Notes Level: This activity can be used in a trigonometry course, assuming students have already had a second year algebra course in high school. Materials: Student pages, calculators Time Frame: Two class periods, with homework. Objectives: To extend students’ knowledge of solving equations by considering a nonalgebraic method. To apply trigonometry to an entirely different field. When to Use: This activity can be used after the students have experience with determining values of trigonometric functions and have a knowledge of basic trigonometric identities. An understanding of other solutions to cubic equations is useful, but not necessary, since this activity does not require prior knowledge of other forms of solution. How to Use: This material can be covered in class by the teacher. In this case, the student pages will serve essentially as textbook material to be read before and after the class presentation. The material could also be assigned for independent or small group work. After completion of the activity, a discussion of the meaning of ”solution of an equation” would be enlightening. Students can consider the question of what kinds of solutions are valid or useful in a given context. √ To start, you might ask about the solutions of the equa2 tion x = 2. Is the solution 2 or is it 1.414 . . . or could it simply be the diagonal of a square of side 1? Answers to Exercises: 1. −1.450, −16.550 2. −1, 5, 4 3. 3.823, −0.893, −2.930 4. −3, 3.303, −0.303

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Solving Cubic Equations Using Trigonometry Student Pages Although by the middle of the sixteenth century Gerolamo Cardano and others had discovered an algebraic formula to solve cubic equations, mathematicians at the time were not comfortable with the use of the formula to solve certain cubics of the form x3 = cx + d, where c and d were positive numbers. The reason was that, in the case where (d/2)2 < (c/3)3 , the formula required the use of complex numbers even though the solutions of the equation were real. The mathematicians felt there should be a way to solve these equations using only real numbers. Here we present one such method, developed around 1600 by Fran¸cois Vi`ete, a French mathematician. This method makes use of trigonometry. We begin with the triple angle formula for the cosine. To derive this, you need to recall the sum formula for the cosine as well as the double angle formula for both the sine and the cosine. cos(α + β) =? sin 2α =? cos 2α =? (There are several ways of writing this double angle formula. You can use whichever one you want.) Now, write cos 3α as cos(2α + α) and expand using the sum formula for the cosine. Simplify the expression. In particular, replace all the occurrences of sin2 α by 1 − cos2 α.

The resulting expression, which only has cosines in it, is cos 3α = 4 cos3 α − 3 cos α. Because we are interested in using this expression to model a cubic equation, we rewrite this in the form 1 3 cos3 α = cos α + cos 3α. 4 4 c 2004, M.A.A.

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We want to use this cubic equation in cos α to solve the cubic equation x3 = cx + d, in the case where (d/2)2 < (c/3)3 . To do this, we set x = r cos α and substitute that in the cubic equation. The result is r3 cos3 α = cr cos α + d or cos3 α =

c d cos α + . r2 r3

If we can turn this equation into our triple angle formula, then the truth of that formula will ensure that x = r cos α is a solution to our equation. To make this transformation, we must have c d 3 1 and cos 3α. = = r2 4 r3 4 q

The first of these equations reduces to 3r2 = 4c or r = 4c/3. The second equation then becomes 1 3d d d q = q . cos 3α = q 3 = 4 (4c/3) 4c/3 4c 4c/3 4c/3 We can rewrite this last result as 3d cos 3α = q . c 4c/3 Thus, if we can find α satisfying this equation, then x q = r cos α satisfies the original cubic. But to be able to find an α, the expression 3d/c 4c/3 must be less than 1 in absolute value. (We do not need to worry about the case where it is exactly 1. Why?) Thus, we require that 9d2 <1 c2 (4c/3)

27d2 <1 4c3

or, finally, that d 2

c 3

c3 d2 < , 4 27

Note that this inequality is precisely the inequality we were given for the relationship between the values c and d in our equation. And, of course, the argument works backwards. If that last inequality is satisfied, the expression for cos 3α is, in fact, less than 1 in absolute value. We can now solve the original cubic equation q using trigonometry. q Simply use the calculator to determine α so that cos 3α = 3d/c 4c/3, and set r = 4c/3. Then it follows that x = r cos α is a solution to the equation x3 = cx + d.

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Examples: q

1. We solve the equation x3 = 300x + 432. In this case, c = 300 and d = 432, so 4c/3 = √ 400 = 20 and cos 3α = 1296/6000 = 216/1000 = 0.216. Using calculators, we find q −1 that cos α = cos((1/3 cos (0.216)) = 0.9. Since r = 4c/3 = 20, we get x = r cos α = 20 · 0.9 = 18 as a solution to the cubic equation. Of course, this is only one solution to the cubic. But once we have found this solution, it is straightforward to find the other two solutions. q

2. Solve x3 = 3x − 1. Here, c = 3 and d = −1, so r = 4c/3 = 2 and cos 3α = −3/6 = −1/2. Then 3α = 120◦ and α = 40◦ . It follows that x = 2 cos 40◦ is one solution to the equation. It is easy enough to find the other two solutions by choosing two other angles where cos 3α = −1/2. If we choose 3α = 240◦ , then α = 80◦ and x = 2 cos 80◦ . If we choose 3α = 480◦ , then α = 160◦ and x = 2 cos 160◦ . We could write the solutions numerically as x = 1.532, x = 0.347, and x = −1.879. Exercises: 1. Find the other two solutions to the cubic equation in example 1. One way to do it is to find additional values whose cosine is 0.216 and then divide by 3. Solve the following cubic equations using this trigonometric method. Find all three solutions in each case. 2. x3 = 21x + 20 3. x3 = 12x + 10 4. x3 = 10x + 3

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Solving Quartic Equations Teacher Notes Level: This activity is designed for use in the second year algebra course in high school. Materials: Student pages Time Frame: This material can be done in one class period. Objectives: To extend students’ knowledge of solving equations by considering the algebraic solution of quartic equations. When to Use: This activity can be used after the students have seen the algebraic solution of cubic equations. How to Use: This material can be covered in class by the teacher. In this case, the student pages will serve essentially as textbook material to be read before and after the class presentation. But the material could also be assigned for independent work or small group work. Answers to Exercises: √ 1. 1 ±

√ 5, −1 ±

√ 2. 1 ± 3.

1 2

√

2, −1 ± 2i 2±

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√ 8 2 + 10 ,

1 2

q √ √ − 2 + 8 2 − 10 i

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Solving Quartic Equations Student Pages Lodovico Ferrari helped Cardano in working through the hints on solving cubic equations provided to Cardano by Tartaglia. But Ferrari also worked out an ingenious method of solving quartic (fourth degree) polynomial equations as well. Cardano provided a sketch of this method in his Ars Magna. Let us assume that the equation has no cubic term and so can be written in the form x4 = cx2 + dx + e. Ferrari’s method is to add second degree terms to both sides to turn each side into a perfect square. To do this, however, requires the knowledge of the solution of cubic equations, which Ferrari, of course, assumes. Once we have the perfect squares, we then take square roots and calculates the answer. In order to follows Ferrari’s method, we need to know the conditions under which the quadratic expression ax2 + bx + c is a perfect square. This happens when the quadratic equation ax2 + bx + c = 0 has precisely one root. And that happens when the expression under the radical sign in the quadratic formula vanishes, that is, when b2 − 4ac = 0. Thus, we have found the necessary condition. When that occurs, the quadratic equation has the single solution x = −(b/2a) and thus ax2 + bx + c factors as 2

ax + bx + c =

√

b ax + √ 2 a

Examples: 1. Solve x4 = 12x − 3. We begin by adding 2zx2 + z 2 to both sides, where z is to be determined. The left side becomes x4 + 2zx2 + z 2 , which is a perfect square. The right side becomes 2zx2 + 12x + z 2 − 3, and we want also to make this into a perfect square. According to our criterion above, this means that 122 − 4(2z)(z 2 − 3) = 0, or 144 − 8z 3 + 24z = 0. Thus our desired value z satisfies the cubic equation 8z 3 − 24z − 144 = 0, or, dividing through by 8, z 3 − 3z − 18 = 0. In theory, we can use the methods already developed to solve this cubic equation. However, in this case, it is clear that one solution is z = 3. We therefore set z = 3 in both the left and right side of our equation. We get x4 + 6x2 + 9 = 6x2 + 12x + 6

(x2 + 3)2 =

√

6x +

√ 2 6 .

If we now take square roots of both sides, we get √ x2 + 3 = ± 6(x + 1). These two equations are quadratic equations in x, which we can solve via the quadratic formula. The solutions to the first equation are x= c 2004, M.A.A.

q √ 1 √ 6± 4 6−6 . 2

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The solutions to the second equation are x=

q √ √ 1 − 6 ± 4 6 + 6i . 2

Thus the given equation has four solutions, two real solutions and two complex solutions. 2. Solve x4 = 10x2 − 4x − 8. To make the left side a square, we add 2zx2 + z 2 ; thus we also add the same quantity to the right side, where z is to be determined. We now have x4 + 2zx2 + z 2 = (10 + 2z)x2 − 4x + z 2 − 8. Thus, the left side is now the square of (x2 + z). The right side will also be a square if (−4)2 − 4(10 + 2z)(z 2 − 8) = 0. We can rewrite this equation as −8z 3 − 40z 2 + 64z + 336 = 0, or, dividing by −8, as z 3 +5z 2 −8z−42 = 0. Again, we could solve this cubic equation by the methods already discussed, but some brief trials shows that z = −3 is a solution. So we set z = −3 in our equation to get x4 − 6x2 + 9 = 4x2 − 4x + 1

(x2 − 3)2 = (2x − 1)2 .

When we take square roots of each side, we get the two quadratic equations x2 − 3 = ±(2x − 1), each of which can be solved by the quadratic formula. The solutions to √ x2 − 3 = 2x − 1, or x2 − 2x − 2 = 0,√are x = 1 ± 3. The solutions to x2 − 3 = −2x + 1, or x2 + 2x − 4 = 0, are x = −1 ± 5. We therefore have four solutions to the original quartic equation, in this case all real numbers. Exercises: Solve the following quartic equations: 1. x4 = 10x2 + 4x − 8 2. x4 = 12x + 5 3. x4 = 6x2 + 8x + 4

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Newtonâ&#x20AC;&#x2122;s Method for Solving Equations Teacher Notes Level: This activity is designed for use in the second year algebra course in high school. Materials: Student pages; calculator Time Frame: This material can be done in one class period, although students should practice the ideas for homework. Objectives: To extend studentsâ&#x20AC;&#x2122; knowledge of solving equations by demonstrating a numerical method for solving equations. When to Use: This activity can be used after the students have seen an algebraic procedure for solving cubic or higher degree polynomial equations. How to Use: This material can be covered in class by the teacher. In this case, the student pages will serve essentially as textbook material to be read before and after the class presentation. But the material could also be assigned for independent work or small group work. Answers to Exercises: 1. 3.04668 2. 1.70998 3. 0.59697 4. 0.75682 5. 0.51800 6. 1.53209, 0.34730, â&#x2C6;&#x2019;1.87939

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Newton’s Method for Solving Equations Student Pages Among Newton’s earliest discoveries was an efficient method for solving polynomial equations numerically. In an earlier activity, we used Newton’s method to solve equations of the form x2 = b, that is, to find square roots. Newton, however, used this method to solve more general polynomial equations. We follow his explanation of this method in terms of the example x3 − 2x − 5 = 0. Newton notes first that the integer 2 can be taken as an initial approximation x1 to a root, because 23 − 2 · 2 − 5 is reasonably close to 0. He then sets x = 2 + p and substitutes in the original equation to get the new equation (2 + p)3 − 2(2 + p) − 5 = 8 + 12p + 6p2 + p3 − 4 − 2p − 5 = p3 + 6p2 + 10p − 1 = 0. Because p is assumed to be small, Newton realized that p2 and p3 are much smaller. Thus, rather than try to solve this cubic equation, he neglects p3 and p2 and solves the resulting linear equation 10p − 1 = 0 to get p = 0.1. It follows that x2 = 2 + 0.1 = 2.1 is the second approximation to the root. We can check that 2.13 − 2 · 2.1 − 5 = 0.061, so x2 is certainly a better approximation to the root than x1 = 2. The next step is to set p = 0.1 + q and substitute that into the equation for p. We get (0.1 + q)3 + 6(0.1 + q)2 + 10(0.1 + q) − 1 = 0, or, 0.001 + 0.03q + 0.3q 2 + q 3 + 0.06 + 1.2q + 6q 2 + 1+10q − 1 = 0. This reduces to q 3 +6.3q 2 +11.23q +0.061 = 0. Again, Newton neglects the terms in q 2 and q 3 . Thus, he reduces the equation to 11.23q + 0.061 = 0, whose solution is q = −0.0054. Therefore, he gets a third approximation for x: x3 = 2.1 − 0.0054 = 2.0946. If we substitute 2.0946 into the polynomial, we get 2.09463 − 2 · 2.0946 − 5 = 0.00005416, and x3 is certainly a much better approximation to the root than x2 . One can continue this method as far as desired. Newton himself stops after one more step with the value x4 = 2.09455148. He notes that if this value is substituted into the original polynomial x3 − 2x − 5, the resulting value is −0.00000001721, certainly close enough to 0 for all practical purposes. In any case, this example shows that Newton’s method gives us a sequence of values which very rapidly approach, or converge to, a solution to the equation. The idea of a sequence of numbers converging to a number is an extremely important one in mathematics. Mathematicians have developed techniques to determine how rapidly a sequence does converge, but we cannot consider those here. Intuitively, however, the fact that two consecutive terms of our approximating sequence agree to four decimal places should convince you that the method does provide a way of approximating the solution very quickly. √ Let us try one more example of Newton’s method. We find 3 2, or, what is the same thing, solve x3 − 2 = 0. Again, we begin with a convenient guess, here x1 = 1. We next set x = 1 + p and substitute this value into the equation. We get (1 + p)3 − 2 = 0, or, 1 + 3p + 3p2 + p3 − 2 = 0, or p3 + 3p2 + 3p − 1 = 0. As before, we neglect the terms in p2 and p2 , so we need to solve 3p − 1 = 0. The solution is p = 0.33333333, so our second approximation is x2 = 1.33333333. We repeat the process by setting p = 1/3 + q and substituting into the equation for p. (Here, it is easier to use a fraction rather than a decimal.) The equation becomes (1/3 + q)3 + 3(1/3 + q)2 + 3(1/3 + q) − 1 = 0. This simplifies to q 3 + 4q 2 + 5 13 q + 10 27 = 0. If c 2004, M.A.A.

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we again ignore terms of degree higher than 1, we get 5 13 q + 10 27 = 0, whose solution is q = 5 − 72 = −0.06944444. Thus, our third approximation to the solution is x3 = 1.33333333 − 0.0694444 = 1.26388889. One more step gives us the approximation x4 = 1.25993349, whose cube is equal to 2.00005924. This approximation is not as accurate as the fourth approximation in our first example. If a better approximation is required, we can simply repeat the process. Newton’s method is an efficient procedure for solving polynomial equations. It does have some minor problems, including the fact that if one is not careful about the initial choice of a first approximation, the procedure may not work at all. But for virtually all reasonable choices, the algorithm is a good one. Despite this, we rarely do hand calculations by Newton’s method or otherwise these days. In general, to solve a polynomial equation approximately, we use the polynomial solving procedure of our calculators. This procedure usually asks for a guess to start with, but then produces an answer accurate to 8 or 9 decimal places in a few seconds. How does it do this? You might write to your calculator manufacturer to determine what algorithm is used. There are several algorithms available for the calculator, including Newton’s method, but different calculator manufacturers may make different choices. Even though you can now solve polynomial equations numerically with a touch of a button or two on your calculator, there is still reason to learn how to solve equations by hand as we have done in this chapter. The electronic solutions do not help you understand the relationships between solutions and coefficients nor do they show you how answers may be expressed in radicals. And, in general, one should not think that the calculator uses “magic” to get a solution. Someone has programmed into the calculator’s innards an algorithm which was originally discovered by someone else hundreds of years ago using hand calculation. So although for quick solutions to polynomial equations in decimals, the calculator is an extremely useful device, it is not the only way you should solve polynomial equations. To appreciate the power of the calculators, it is useful to go through, as we have, the struggles of people over the centuries to learn how to use various methods to find solutions. Exercises: Use Newton’s method to find one solution to each of the following five equations. Begin with the value given. Check the solution by using your calculator. 1. x3 − 6x − 10 = 0 (Begin with x1 = 3.) 2. x3 − 5 = 0 (Begin with x1 = 2.) 3. 2x3 − 4x2 + 1 = 0 (Begin with x1 = 1.) 4. x3 + x2 − 1 = 0 (Begin with x1 = 1.) 5. x4 − 4x + 2 = 0 (Begin with x1 = 12 . What happens if you begin with x1 = 1?) 6. Find all three solutions to x3 − 3x + 1 = 0. You will need to pick a different initial value in each case. c 2004, M.A.A.

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The Relationship Between Roots and Coefficients of a Polynomial Teacher Notes Level: This activity is designed for use in the second year algebra course in high school. Materials: Historical background; student pages Time Frame: Each of the three parts can be done in one class period. Objectives: To extend students’ understanding of polynomials by looking at the relationship between the roots and the coefficients of a polynomial. When to Use: The first part of this activity can be used after the students have seen the algebraic solution of quadratic equations. Although it is not essential, it would be useful for the second and third parts for the students to have considered the solutions of cubic and quartic equations. How to Use: This material is designed for small group work. There could be a whole class discussion after each group has worked through the material. Answers to Exercises: Exercise Set 1

2. −3, − 32 ±

1. x2 + x − 6 = 0

3. 2, 3, −3

2. x2 − 2x − 35 = 0

4. 3, 3, −1

3. x2 − 6x + 7 = 0

5. −5, 7, −3

4. x2 + 10x + 25 = 0

√ 1. 1 ±

6. x2 − (r + s)x + rs = 0

2. 1 ±

1. 3,

− 52

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1 2

√ 5

Exercise Set 3

5. x2 − 4x + 5 = 0

Exercise Set 2

1 2

2, −1 ± 2i √

√

1 2 1 2

√

√ 5, −1 ±

2±

√ 8 2 + 10 ,

q √ √ − 2 ± 8 2 − 10 i

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The Relationship Between Roots and Coefficients of a Polynomial Student Pages Historical Background Although Gerolamo Cardano seemed to have some limited understanding of the relationship between the roots and the coefficients of cubic polynomials, he could not deal with a general rule because he was still uncomfortable with negative roots and certainly could not understand complex roots. Fran¸cois Vi`ete (1540–1603), in his work on equations around 1600, noted the relationship between roots and coefficients in polynomials up to degree five, but did not give any proof of the result. And, in general, he also only dealt with positive roots of polynomials. Probably in the second decade of the seventeenth century, Thomas Harriot (1560– 1621) was able to work out the relationship between roots and coefficients in some detail, at least for equations of degree up to four. Harriot was willing to use negative and complex roots, and so used this relationship to aid in solving equations. But Harriot’s work was not published during his lifetime; it was only left in manuscript form to be shared among some of his friends. Thus it was Albert Girard (1595–1632) who was the first to publish a detailed study of this relationship in his 1629 work, A New Discovery in Algebra. Not only did he state this relationship in detail, but he also asserted the truth of the fundamental theorem of algebra, that every polynomial equation has as many solutions as its degree. He was not able to prove these results, but he certainly was able to use them. In particular, he understood that it was absolutely necessary to deal with negative and complex roots of polynomials to allow him to state the general rules.

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Relating Roots and Coefficients of Quadratic Polynomials Once mathematicians were comfortable with negative roots, it was probably very easy to discover the relationship between the roots and the coefficients of a quadratic polynomial, especially one whose initial coefficient was 1. You can discover this relationship as well. So, for each of the following quadratic polynomials, find the two roots (either by using the quadratic formula or by some other method). Then find the sum of the solutions and the product of the solutions. In each case, compare your answers to the coefficients of the polynomial, that is, to the values b and c in the polynomial x2 + bx + c. Polynomial

Solutions

Sum of Solutions

Product of Solutions

x2 − 5x + 6 x2 − 10x − 144 x2 + 5x + 6 x2 − 6x + 8 x2 − 12 x − 14 x2 − 3x − 3 x2 −

√ 18x − 4 12

Make a conjecture about the relationship of the roots to the coefficients. The conjecture should be of the form “The sum of the roots of the quadratic polynomial x2 +bx+c is equal to and the product of the roots of the polynomial is equal to ,” where the blanks are filled in with expressions involving b and/or c. Next, recall that the quadratic formula gives the two roots of x2 + bx + c as √ b x1 = − + 2

b2 − 4c 2

√ and

b x2 = − − 2

b2 − 4c . 2

Add x1 to x2 to confirm your conjecture about the sum of the roots; then multiply x1 by x2 to confirm your conjecture about the product of the roots. In each of the above examples, there were two distinct roots of the polynomial. How can you modify your conjecture to take care of the case where there is only one root of the polynomial? In answering this question, consider, for example, the polynomials x2 + 4x + 4 and x2 − 6x + 9. Also, in each of the above examples, the roots of the polynomial are real numbers. If you understand that the roots may be complex, show that this makes no difference in your conjecture. That is, the same rule applies even if the two roots of the polynomial are complex numbers. c 2004, M.A.A.

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Exercise Set 1: Using your conjecture, which you have now turned into a theorem by proving it algebraically, answer the following: 1. Find a quadratic equation whose two solutions are 2 and −3. 2. Find a quadratic equation whose two solutions are −5 and 7. √ √ 3. Find a quadratic equation whose two solutions are 3 + 2 and 3 − 2. 4. Find a quadratic equation which has the single solution −5. 5. Find a quadratic equation whose two solutions are 2 + i and 2 − i. 6. Find a quadratic equation whose two solutions are r and s, where r and s can stand for any numbers, real or complex.

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Relating Roots and Coefficients of Cubic Polynomials You have determined above that the sum of the two roots of the quadratic polynomial x2 + bx + c = 0 is −b, the negative of the coefficient of x, while the product of the two roots is c, the constant term. We can demonstrate this in a slightly different manner than above, where you used the quadratic formula and checked. We start with two arbitrary solutions to a quadratic equation, x = r and x = s. We can rewrite these solutions as the equations x − r = 0 and x − s = 0. We then multiply these two equations together. We get (x − r)(x − s) = 0, or x2 − (r + s)x + rs = 0. This equation has precisely the two roots x = r and x = s. Further, we see by looking at the form of the equation that, as promised, the constant term is the product of the roots, while the sum of the solutions is the negative of the coefficient of x. We now want to extend this idea to cubic polynomials. So let us assume we have three solutions to a cubic, x = r, x = s, and x = t. We can rewrite these expressions in the form of three equations x − r = 0, x − s = 0, and x − t = 0. We then multiply these three together. As above, the product of the first two equations is the equation x2 − (r + s)x + rs = 0. So let us multiply that equation by x − t = 0. We get (x2 − (r + s)x + rs)(x − t) = x3 − (r + s)x2 + rsx− tx2 + t(r + s)x − rst = x3 − (r + s + t)x2 +(rs+ rt + st)x − rst = 0. That is, if r, s, t are the solutions of a cubic equation, then the equation may be written in the form x3 − (r + s + t)x2 + (rs + rt + st)x − rst = 0. Using this equation, we see how the coefficients of the cubic polynomial are related to the three solutions, in a way analogous to how we did this for quadratic equations. The constant term in the equation is the negative of the product of the solutions. The coefficient of x2 is the negative of the sum of the solutions. And the coefficient of x is the sum of all products of pairs of the solutions. As an example of the relationship, let us consider the equation x3 − 9x2 + 26x − 24 = 0, whose solutions are x = 2, x = 3, and x = 4 (check these). In this case, the constant terms of the equation is −24, which is the negative of the product 2 · 3 · 4 of the solutions. Also, the coefficient of x2 is −9, which is the negative of the sum 2 + 3 + 4 of the solutions. Finally, the coefficient of x is 26, which is the sum of the products of all pairs of the solutions: 2 · 3 + 2 · 4 + 3 · 4 = 26. Now in the case of the quadratic equation, we can use the relationship between the solutions and the coefficients to determine the solutions. For we can often find by inspection two numbers whose sum and product are given. It is in general much more difficult to use these relationships to solve a cubic. If we have a cubic equation of the form x3 + bx2 + cx + d = 0, it may be quite difficult to find three numbers whose product is −d, whose sum is −b, and such that the sum of all products of pairs is c. On the other hand, we can search for a single solution. After all, the constant term is the negative of the product of the solutions. Put the other way around, any solution must be a factor of that constant term. In particular, assuming we are dealing with a polynomial with integral coefficients, any integral solution must be a factor of the constant c 2004, M.A.A.

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term. Since this term has only a finite number of factors, we can easily check all of them to determine if one of them is a solution. And once we have one solution, we can use the coefficient-root relationship to find the others. Examples: 1. Given the equation x3 − 15x − 4 = 0, we note that since the constant term is −4, its only factors are ±1, ±2, and ±4. So there are only six possible integral solutions to the equation. Trying each of these six values in turn, we find that indeed 4 is a solution. But now that we have one solution, we can find the others. Because the product of all three solutions is 4 and one solution is 4, the product of the other two solutions must be 1. Also, because the sum of all three solutions is 0, and one solution is 4, the sum of the other two solutions must be −4. Thus, we need to find two numbers whose product is 1 and whose sum is −4. But these √ two are the roots of the quadratic equa−2 ± 3. It follows tion x2 + 4x + 1 = 0, namely, x = √ √ that the three solutions of the original cubic are x = 4, x = −2 + 3, and x = −2 − 3. 2. In this example, you should determine the appropriate numbers. So take x3 + 6x2 + x − 14 = 0. The factors of −14, and therefore the possible integral solutions, are ±1, ±2, ±7, and ±14. We try each in turn and discover that −2 is a solution. Therefore, the product of the other solutions must be , while the sum of the other solutions must be . These two additional solutions must then be solutions of the quadratic equation . The quadratic formula gives the solution x= . So we again have our three solutions: x = −2, x = , and x = . 3. For a final example, we try x3 − 27x + 54 = 0. There are many factors of 54, and therefore many possible integral solutions to the equation. With patience, we find that x = 3 is a solution. So we now know that the product of the other two solutions is while their sum is . But here we can find numbers by inspection, namely and . It follows that the original cubic only has two distinct solutions, x = and x = . However, in order that our relationship between roots and coefficients holds, we must consider x = 3 as a double solution, a solution which we count twice. Recall that for the analogous relationship to hold for quadratic equations, in an equation with a single root, we must also count that root twice. Exercise Set 2: Solve each of the following cubic equations. First, find one solution r by trying the factors of the constant term. Next, find the relationship the other two solutions must satisfy and thus determine the quadratic equation of which they are roots. Finally, find the roots of the quadratic equation. 1. x3 + 2x2 − 13x − 6 = 0

4. x3 − 5x2 + 3x + 9 = 0

2. x3 + 6x2 + 10x + 3 = 0

5. x3 + x2 − 41x − 105 = 0

3. x3 − 2x2 − 9x + 18 = 0 c 2004, M.A.A.

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Relating Roots and Coefficients of Quartic Polynomials We can apply the same procedure to fourth degree polynomials that we applied to cubics. Namely, let us assume that there are four roots of the fourth degree polynomial, x = r, x = s, x = t, and x = u. As before, we rewrite these results in the form of equations: x − r = 0, x − s = 0, x − t = 0, and x − u = 0. We then multiply these equations together: (x − r)(x − s)(x − t)(x− u) = 0

(x3 − (r + s + t)x2 +(rs+ rt + st)x − rst)(x− u) = 0.

With a little further manipulation, which you should check carefully, we finally get x4 − (r + s + t + u)x3 + (rs + rt + ru + st + su + tu)x2 − (rst + rsu + rtu + stu)x + rstu = 0. In words, the constant term of the equation is the product of the four roots, the coefficient of x is the negative of the sum of all products of sets of three roots, the coefficient of x2 is the sum of all products of sets of two roots, while the coefficient of x3 is the negative of the sum of the roots. For example, by multiplying together the four equations x − 2 = 0, x − 3 = 0, x − 4 = 0, and x + 5 = 0, we get the equation x4 − 4x3 − 19x2 + 106x − 120 = 0. The product of the four roots, namely 2 · 3 · 4 · (−5) is −120 as our relationship shows. If we add together all the products of sets of three of these roots, we get 24 − 30 − 40 − 60 = −106, the negative of the coefficient of x. Also, the sum of all the products of pairs of these roots is 6 + 8 − 10 + 12 − 15 − 20 = −19, the coefficient of x2 . Finally, the sum of the roots is 2 + 3 + 4 − 5 = 4, the negative of the coefficient of x3 . Although we can attempt to solve fourth degree equations by a method analogous to the method used previously to solve cubics, it is frequently a very inefficient method. But Louis Lagrange, around 1770, found a somewhat different method for solving fourth degree polynomials using the coefficient-root relationship. We sketch the method here. We want to solve the fourth degree polynomial equation x4 + ax3 + bx2 + cx + d = 0. The first step is to use a linear substitution to remove the 3rd degree term. Namely, we set x = y − a4 and substitute into the equation. That is, we rewrite the equation in the form a 4 a 3 a 2 a y− +a y− +b y− +c y− + d = 0. 4 4 4 4 Show that when you expand this, the two terms in x3 cancel each other out. Thus we are left with an equation of the form y 4 + βy 2 + γy + δ = 0. (It is not necessary to work out here what β, γ, and δ are in terms of the original coefficients. In any given case, one just does the substitution and figures out these new coefficients.) We will now solve the new equation for y. After doing so, we simply subtract a4 from each solution to get the corresponding values for x. To solve the equation, we will make c 2004, M.A.A.

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use of the relationship between the roots r, s, t, u of this new equation and the corresponding coefficients β, γ, and δ. Thus r + s + t + u = 0, rs + rt + ru + st + su + tu = β, rst + rsu + rtu + stu = −γ, and rstu = δ. But we will use a clever idea of Lagrange. Namely, set A = rs + tu, B = rt + su, and C = ru + st. The idea is to find a cubic equation satisfied by A, B, and C. Because we already know how to solve a cubic, we will then use the values of A, B, and C to find the roots r, s, t, and u. To find the cubic equation satisfied by A, B, and C, we need to find their sum, the sum of all products of pairs, and their product. First, we note that A + B + C = β. Second, we calculate AB + AC + BC. Show by multiplying and simplifying that this sum of products equals −4δ.

Finally, we calculate ABC. Although the calculation is very messy, show that ABC = γ 2 − 4βδ.

Therefore, we can use the relationship between coefficients and roots of cubic equations to show that A, B, and C are solutions of the cubic equation z 3 − βz 2 − 4δz − (γ 2 − 4βδ) = 0. Given that we know how to solve a cubic, either by Cardano’s formula or by checking values which divide the constant term or by some other method, we can assume that A, B, and C are now known. We will use the value of A to find the values of r, s, t, and u c 2004, M.A.A.

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by solving two quadratic equations. First we find the quadratic equation whose roots are rs and tu. To do this, we need to find their sum and product. But from what we already know, we have that rs + tu = A and that (rs)(tu) = δ. Therefore, the quadratic equation satisfied by µ = rs and ν = tu is . (Fill in the equation from what you know about quadratic equations with given roots.) We can, of course, solve this equation, but for now we will just assume that the roots are µ and ν. Since µ is the product of r and s, in order to find these values, we need to know their sum. To calculate that, we note that (r + s) + (t + u) = 0

(why?)

and tu(r + s) + rs(t + u) = −γ

(why?)

If we now solve the first equation for t + u and substitute in the second, we get tu(r + s) − rs(r + s) = −γ

(tu − rs)(t + u) = −γ

(µ − ν)(t + u) = γ.

We can now solve for r + s and also for t + u = −(r + s). We get γ γ r+s= =ρ and t+u=− = σ. µ−ν µ−ν Because we now know the sum and product of r and s, we can find the quadratic equation that r and s satisfy. That equation is . Similarly, because we know the sum and product of t and u was can also find the quadratic equation that t and u satisfy. We can now solve each of these quadratic equations to find the four roots of our equation for y. Recall that the four solutions for x come from subtracting a/4 from the values for y. Example: We solve x4 − 12x + 3 = 0. Because there is already no x3 term, we do not have to do the linear substitution. We therefore just use this equation as it is to set β = 0, γ = −12, and δ = 3. It follows that A, B, and C are roots of the cubic equation z 3 − 12z − 144 = 0. To solve this cubic, it is easiest to note that an integral solution must be a divisor of 144. Although there are many such divisors, some trials will show that one such solution is A = 6. (As our discussion above shows, it is not necessary to determine B and C.) Given that we know A, we see that the quadratic equation satisfied and √ by µ = rs √ 2 ν = tu is w − 6w + 3 = 0. The roots of this equation are µ = 3 + 6 and ν = 3 − 6. We now calculate √ √ γ −12 γ r+s= = √ =− 6=ρ = 6 = σ. and t+u = − µ−ν µ−ν 2 6 √ √ Therefore, the quadratic equation satisfied by r√and s is v 2 + 6v + 3 + 6 = 0 and the √ equation satisfied by t and u is v 2 − 6v + 3 − 6 = 0. Although these quadratic equations are a bit messy, a straightforward use of the quadratic formula gives us √ √ q q √ √ 6 1 6 1 r=− + + −6 − 4 6 = − 6 + 4 6i. 2 2 2 2 c 2004, M.A.A.

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Similarly,

√ q √ 6 1 − 6 + 4 6i, s=− 2 2 √ q 6 1 √ + 4 6 − 6, t= 2 2

and

√ q 6 1 √ u= − 4 6 − 6. 2 2 We have now completely solved the fourth degree equation. Note that two of the roots are real (t and u), while the other two are complex (r and s). One can check that the sum of the four roots is 0, the coefficient of x3 in the original equation, as expected. It is a bit trickier to check that the product of the four roots is 3, but you should try that. Exercise Set 3: Solve the following quartic equations using Lagrange’s method: 1. x4 − 12x − 5 = 0 2. x4 − 10x2 − 4x + 8 = 0 3. x4 − 6x2 − 8x − 4 = 0

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Finding Maxima and Minima of Polynomials Teacher Notes Level: This activity is designed for use in a precalculus course. Materials: Historical background; student pages; graph paper Time Frame: This material can be done in two class periods with homework. Objectives: To extend students’ understanding of polynomials by looking at the maxima and minima in their graphs. When to Use: This activity can be used once students understand basic graphing techniques and have some feeling for the types of graphs determined by polynomial functions. How to Use: This material can be covered in class by the teacher. In this case, the student pages will serve essentially as textbook material to be read before and after the class presentation. But the material could also be assigned for independent work or small group work. Answers to Exercises: q

2. x = ± b/3; the positive value gives the maximum. b 3. Maximum or minimum occurs at x = − 2a .

4. x = ±1 6. x = 0, x = 1, and x = −1.

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Finding Maxima and Minima of Polynomials Student Pages Historical Background Mathematicians have sought to find the largest (or smallest) values of a particular function for millennia. One simple problem, whose roots are in ancient Greece, is to find the point at which to divide a line of length b so that the rectangle made from the two pieces has the largest possible area. A related three-dimensional version might be to find the cylindrical can of largest volume which can be made from a given amount of material. Or we might want to find the rectangle of least perimeter which surrounds a given area. Although ad hoc techniques for solving specific maximum or minimum problems were worked out in ancient Greece and in medieval Islam, it was not until the seventeenth century that general methods for developed. The person responsible for the first of these general methods was Pierre de Fermat (1601â&#x20AC;&#x201C;1665), a French lawyer who did mathematics in his spare time. Although he did not write any textbooks, or even publish any articles, his work was circulated in manuscripts throughout France and the rest of Europe. Among his accomplishments were the invention of analytic geometry and significant contributions to the solution of the two problems eventually forming the calculus: the problem of extrema and the problem of finding areas. In this activity, we consider his solution to the problem of finding maxima and minima.

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Fermat’s Method of Finding Extrema Recall that in the quadratic polynomial x2 + bx + c, the coefficient b is equal to the negative sum of the two roots, while the constant term c is equal to the product of the two roots. We know this relationship holds whether the roots are positive, negative, or even complex. However, in the seventeenth century, mathematicians were not too comfortable with either negative or complex roots of equations. Thus, Fran¸cois Vi`ete, for example, only considered quadratic equations which had two positive roots in dealing with the relationship between the roots and the coefficients. Such a quadratic equation can be written in the form bx − x2 = c. To discover the relationship between the two roots, x1 and x2 , he equated the two expressions bx1 − x21 and bx2 − x22 . Then x21 − x22 = bx1 − bx2

(x1 + x2 )(x1 − x2 = b(x1 − x2 .

Thus, when he divided through by x1 − x2 , he found that x1 + x2 = b, that is, “b is the sum of the two roots being sought.” Substituting x1 + x2 for b in the equation bx1 − x21 = c, he found the other relationship x1 x2 = c, or “c is the product of the two roots being sought.” Fermat, in the late 1620s, was stimulated to consider the question of extrema by a study of Vi`ete’s work relating the coefficients to the roots of a polynomial. Let us consider the problem of dividing a line of length 10 into two parts whose product is maximal. In other words, we want to maximize the function x(10 − x) or 10x − x2 . Fermat knew from Euclid that the maximum possible value of this function was 25 = (10/2)2 and also that for any number less than the maximum, there were two possible values for x whose sum was 10. But what happened as the function approached its maximum value? The geometrical situation made it clear to Fermat that even for this maximum value, the equation 10 had two solutions, each of the same value: x1 = 10 2 = 5 and x2 = 10 − x1 = 10 − 2 = 10 ete, he set 10x1 − x21 = 10x2 − x22 . He rearranged this as 2 = 5. Thus, following Vi` 10x1 − 10x2 = x21 − x22 , or as 10(x1 − x2 ) = (x1 + x2 )(x1 − x2). Dividing by x1 − x2 , he derived the same result as Vi`ete: 10 = x1 + x2 , an equation holding for any two roots. Since the maximum occurs when the two roots are identical, he set x1 = x2 (= x) and found that 10 = 2x. Thus the maximum occurs when x = 10 2 = 5. In general, Fermat could find the maximum for the function bx − x2 by the same technique. Setting bx1 − x21 = bx2 − x22 , he could rearrange and then divide by x1 − x2 . Since now b = x1 + x2 , he could set x1 = x2 = x and get b = 2x. Then the maximum occurs when x = 2b . This insight gave Fermat a general method for maximizing a polynomial p(x): Set p(x1 ) = p(x2 ). Then divide through by x1 − x2 to find the relationship between the coefficients and any two roots of the polynomial. Finally, set the two roots equal to one another and solve. Thus, to maximize bx2 − x3 , Fermat set bx21 − x31 = bx22 − x32 and derived b(x21 − x22 ) = x31 − x32

b(x1 − x2 )(x1 + x2 ) = (x1 − x2 )(x21 + x1 x2 + x22 ).

Then bx1 + bx2 = x21 + x1 x2 + x22 . Fermat then set x1 = x2 (= x) and determined that 2bx = 3x2 , from which he concluded that the maximum value occurs when x = 2b 3 . In c 2004, M.A.A.

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fact, Fermat realized that in general his method would give him either a maximum of a minimum for the function. So to decide which of the two cases held, he used geometry. After all, his problems were algebraic translations of geometric problems. Fermat’s method does raise a significant question. How can one divide through by x1 − x2 and then set that value equal to 0? After all, you know that you are not allowed to divide by 0. For Fermat, the geometric situation showed that the roots were even distinguishable when their difference was 0. Thus, he never felt he was dividing by 0. He simply assumed that the relationships worked out using Vi`ete’s methods were perfectly general (for example, x1 + x2 = b) and thus held for any particular values of the variables, even those at the maximum. Fermat did realize, however, that if the polynomial p(x) were somewhat complicated, the division by x1 − x2 might be rather difficult. Thus he modified his method to avoid this. Instead of considering the two roots as x1 and x2 , he wrote them as x and x + e. Then, after equating p(x) with p(x + e), he had only to divide by e or one of its powers. In the resulting expression he then removed any term that contained e to get an equation enabling the maximum to be found. Thus, using his original example of p(x) = bx − x2 , Fermat put bx − x2 equal to b(x + e) − (x + e)2 = bx − x2 + be − 2ex − e2. Canceling common terms gave him be = 2ex + e2 . After dividing by e, he found b = 2x + e. Removing the term which contains e gave Fermat his known result: x = 2b . Exercises: 1. Use Fermat’s second method to do the second example: Find x such that bx2 − x3 reaches a maximum value. 2. Use either one of Fermat’s methods to find x which maximizes bx − x3 . Note that Fermat’s method gives you two answers here. How do you decide which is one provides the maximum? What can you say about the second answer? 3. Use either of Fermat’s methods to find the value of x which gives the maximum or minimum of ax2 + bx + c. How do you know whether you have found a maximum or a minimum? Will you always find one of these? Can a quadratic function have more than one maximum or minimum? 4. Use either of Fermat’s methods to find where the maximum and/or minimum of x3 − 3x + 2 occurs. You should find two different answers. How do you know which is a maximum or which is a minimum, or is it possible that both are maximums or both minimums? 5. How many maximums or minimums can a cubic function have? Does a cubic always have a maximum or a minimum? Consider, for example, the function x3 + 3x and apply Fermat’s method. What happens here? 6. Use either of Fermat’s methods to find where the maximum and/or minimum of x4 − 2x2 + 1 occurs. How many maximums are there? How many minimums are there? What degree equation must you solve to answer these questions? c 2004, M.A.A.

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7. How many maximums or minimums can a fourth degree polynomial function have? Does a fourth degree polynomial always have a maximum or a minimum? How do you know? 8. Generalize the results of exercises 3, 5, and 7 to determine the number of maximums or minimums a polynomial of degree n can have. Does an nth degree polynomial always have a maximum or a minimum? Consider whether n is positive or negative.

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Bibliography Berggren, J. Lennart (1986). Episodes in the Mathematics of Medieval Islam. New York: Springer Verlag. Cardano, Gerolamo (1968). The Great Art, or The Rules of Algebra. Trans. by T. Richard Witmer. Cambridge: MIT Press. Cuomo, Serafina (2001). Ancient Mathematics. London: Routledge. Descartes, René (1954). The Geometry of René Descartes. Trans. by Eugene Smith and Marcia Latham. New York: Dover. Heath, Thomas (1921). A History of Greek Mathematics. New York: Dover (1981 reprint edition). Joseph, George Ghevergese (2000). The Crest of the Peacock: Non-European Roots of Mathematics. Princeton: Princeton University Press. Katz, Victor J. (2004). A History of Mathematics. Boston: Addison-Wesley. Li Yan and Du Shiran (1987). Chinese Mathematics: A Concise History. Trans. by John Crossley and Anthony W.-C. Lun. Oxford: Oxford University Press. Martzloff, Jean-Claude (1997). A History of Chinese Mathematics. Trans. by Stephen Wilson. Berlin: Springer Verlag. Neugebauer, Otto (1951). The Exact Sciences in Antiquity. Princeton: Princeton University Press. Shen Hangshen, John Crossley, and W.-C. Lun (1999). The Nine Chapters on the Mathematical Art: Companion and Commentary. Oxford: Oxford University Press. Van der Waerden, B. L. (1961). Science Awakening I. New York: Oxford University Press.

Web Resources British Society for the History of Mathematics: http://www.dcs.warwick.ac.uk/bshm/resources.html David Joyce’s History of Mathematics Home Page: http://aleph0.clarku.edu/~djoyce/mathhist/ St. Andrews MacTutor History of Mathematics Archive: http://www-history.mcs.st-and.ac.uk/history/

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