1. Equations and Functions
Algebraic methods for solving equations

2. Index Introduction to algebraic equations Quadratic equation
Complete quadratic equation
Incomplete quadratic equation
Solving Incomplete quadratic equations
The quadratic formula Sridhara (Bhaskara)
Solving complete quadratic equations
Equations bi-quadratic
Del Ferro, tartaglia and cardano’s history
General cubic equation (solving cubic equations)

3. Introduction to algebraic equations
Algebraic equations are equations where the unknown x is subject to algebraic operations such as addition, subtraction, multiplication, division
examples: to x + b = 0 to x ² + bx + c = 0 at x4 x ² + b + c = 0

4. Quadratic equation
A quadratic equation in the unknown x is of the form: to x ² + bx + c = 0 where the real numbers a, b and c are the coefficients of the equation, and that must be nonzero. This equation is also called a quadratic equation, because the highest degree term is squared.

5. Complete quadratic equation
A quadratic equation is complete, if all the coefficients a, b and c are nonzero. examples: 2 x ² + 7x + 5 = 0 3 x ² + x + 2 = 0 Incomplete quadratic equation A quadratic equation is incomplete if b = 0 or c = 0 or b = c = 0. Incomplete in the equation the coefficient is nonzero. examples: 4 x ² + 6x = 0 3 x ² + 9 = 0 2 x ² = 0

6. Solving Incomplete quadratic equations Equations of the type ax ² = 0: Divide the whole equation for obtain: x ² = 0 meaning that the equation has two roots equal to zero. Equations of the type ax ² + c = 0: Divide the whole equation by a and pass the constant term to the second member to get: x ² =-c / a If-c / a is negative, there is no solution in real numbers. If-c / a is positive, the equation has two roots with the same absolute value (modulus) but of opposite signs. Equations of the type ax ² + bx = 0: Factored to obtain the equation: x (ax + b) = 0 and the equation has two roots: x '= 0 or x‘’ =-b / a

7. The quadratic formula Sridhara (Bhaskara)
We will show in sequence as the mathematician Sridhara, obtained the formula (known as) of Bhaskara, which is the general formula for solving quadratic equations. A curious fact is that the formula of Bhaskara was not discovered by him but by the Indian mathematician Sridhara at least a century before the publication of Bhaskara, a fact recognized by the Bhaskara, although the material has not built by the pioneer reached us. The basis used to obtain this formula was to find a way to reduce the quadratic equation to one of the first degree, by extracting square roots of both members of the same.

8. Solving complete quadratic equations Bhaskara formula:
D = b ²-4ac
x ² - 5 x + 6 = 0  Identifying the coefficients: a = 1, b = -5, c = 6 Write D = b ²-4ac. Calculate D = (-5) ² -4 × 1 × 6 = = 1
Replace the values ​​of the coefficients a, b and c in the formula v

9. Equations bi-quadratic equations bi- quadratic are equations of the fourth degree in the unknown x, the general form: at x4 x ² + b + c = 0 In fact, this is an equation that can be written as a quadratic equation by replacing: y = x ² to generate a ² + y b y + c = 0 We apply the quadratic formula to solve this last equation and obtain solutions y 'and y "and the final procedure should be more careful, since x ² = y ² = x or y " and y 'or y "is negative, the solutions do not exist for x.

10. Del Ferro, tartaglia and cardano’s history
Del Ferro, Tartaglia and Cardano, provided a tragicomedy of disputes, achievements and disappointments.
The first to develop a mathematical method for solving cubic equations of the form x3 + ax + b = 0 was Scipione del Ferro, teacher at the University of Bologna, Italy, in the passage of the 15th century to 16th century. Before death, he revealed his method, which had kept secret, the Antonio Fiore.
Nicoll Tartaglia was born in Brescia, Italy in It is said that he was so poor as a child that studied mathematics writing on the tombstones of a cemetery. In 1535 he was challenged by Antonio Fiore to a mathematical competition. At the time, academic disputes were common, often rewarding the winner with the use of the loser. Tartaglia could solve the cubic equations of DelFerro, but had also discovered a method to solve the cubic form x3 + ax2 + b = 0 Armed with this knowledge, was the winner in the competition.

11. Tartaglia's last years were embittered by a quarrel with Girolamo Cardano ( ), an Italian mathematician who, besides the famous doctor in Milan, was also an astronomer. In 1570, Cardano was arrested for heresy, for having written a horoscope of Jesus Christ. In 1539, at his home in Milan, Cardano persuaded Tartaglia to tell him his secret method for solution of cubic equations, under oath never to divulge it.
Some years later, however, Cardano learned that part of the method consisted of a posthumous publication of Del Ferro. He then decided to publish a complete study of cubic equations in his treatise Ars Magna (1545), a work that surpassed all algebra books published so far. In Ars Magna Cardano exposes a method to solve the cubic equation based on geometric arguments. There also exposes the general solution of the quartic equation AX4 BX3 cx2 dx = 0, and discovered by Ludovico Ferrari ( ), pupil of Cardano, who seems to have surpassed the master in the algebra of polynomial equations.

12. In 1548, Tartaglia challenged Cardano for a math competition to be held in Milan did not attend Cardano, Ferrari and sent to represent him. It seems that Ferrari won the race, which caused the death Tartaglia unemployment and poverty in the nine years later.

13. General cubic equation
The general cubic equation is A x3 + B x2 + C x + D = 0 The coefficients A, B, C, D are real or complex numbers with A not 0. Dividing through by A, the equation comes to the form x3 + b x2 + c x + d = 0

14. The quadratic term disappears Now we want to reduce the last equation by the substitution x = y + r The cubic equation becomes: (y + r)3 + b (y + r)2 + c (y + r) + d = 0 <=> y3 + (3 r + b) y2 + (3 r2 + 2 r b + c) y + r3 + r2 b + r c + d = 0 Now we choose y such that the quadratic term disappears choose r = -b/3

15. So, with the substitution
x = y – b/3
the equation
x3 + b x2 + c x + d = 0
comes in the form
y3 + e y + f = 0

16. Vieta's substitution To reduce the last equation we use the Vieta substitution y = z + s.1/z The constant s is an undefined constant for the moment. The equation y3 + e y + f = 0 becomes (z + s/z )3 + e (z + (s/z)) + f = 0 expanding and multiplying through by z3 , we have z6 + (3 s + e) z4 + f z3 + s (3 s + e) z2 + s3 = 0 Now we choose s = -e/3. The equation becomes z6 + f z3 - e3/27 = 0 With z3 = u u2 + f u -e3/27 = 0 This is an easy to solve quadratic equation.

20. http://en. wikipedia. org/wiki/Equation http://en. wikipedia

22. ESDMI 11ºA Pedro Batista, nº 19