History of algebra 2

Author: Matthew David

History of algebra 2

Introduction to history of algebra 2:

Algebra is a branch of mathematics. Algebra plays an important role in our day to day life. Algebra 2 covers the four basic operations in algebra such as addition, subtraction, multiplication and division. The most important terms variables, constant, coefficients, exponents, terms and expressions are explained in algebra 2. We will know the symbols and alphabets in the place unknown a value by the help of algebra 2. Therefore, students are getting algebra 2 for their studies. But in this article we are going to see the history of algebra.

Type in algebra 2 questions:

In this article we discuss type in algebra 2 questions solving. In algebra 2 questions solving are easy to understand and solve. Algebra is one of the main branches of arithmetic. It explains the interaction and properties of quantity by means of letters and other signs. The basic algebra has the following subtopics are

Variables,
Expressions,
Terms,
Polynomials,
Equations

The solving type in algebra 2 questions are given below.

History of algebra 2:

The History of geometric algebra was invented by Greeks. The Greek is the father of the algebra. He worked with algebraic equations.

The History of word algebra was derived the Arabic language. The most of the algebraic method are invented by Arabic mathematician Muhammad ibn Musa al-Khwarizmi. He studied Indian mathematics addition and subtraction and introduced these operations (addition and cancellation) in algebra.

Al-Khowarizmi used the words jabr and muqubalah to point out the basic operation of the equation. The word Jabr represent the subtraction of the both sides and the word nuqubalah represented the like terms to cancel."

The solution for the cubic equation is discovered by mathematical inventor by omer. Italian mathematician and Leonardo Fibonacci proved the history of the cubic equation  by the approximated value.

The Italian mathematicians Scipione del Ferro, Niccolo Tartaglia and Gerolamo got the exact solution for general cubic equation by using constant term.

Cardano's pupil, Ludovico Ferrari were got the solution fourth degree equation.

In algebra, symbols are used which was introduced in early 16th century.

The French mathematician Rene invented analytic geometry, which reduces the steps to solve geometric and algebraic.

In 18th century, the German mathematician Carl Friedrich Gauss had proven that every polynomial equation must have at least one root in the complex. Due to this invention, algebra had got new phase. Therefore, the concentration moved to polynomial equation.

British mathematician William Rowan Hamilton discovered the History of Quaternion. He extended the arithmetic complex numbers.

Type in algebra 2 questions:

Example 1:

Solve 5x – 6 = 3x – 12

Solution:

Given expression 5x – 6 = 3x -12

Add 6 on both sides of the equation

5x -6 + 6 = 3x – 12 +6

5x = 3x -6

Subtract 3x on both sides of the equation

5x – 3x = 3x – 3x – 6

2x = -6

Divide 2 on both sides of the equation

`(2x)/2` = `-6/2`

x = `-6/2`

x = -3

Solution is x = -3

Example 2:

Solve the equation: x+2y+3z =14; 3x+y+2z = 11; 2x+3y+z = 11.

Solution:

x+2y+3z = 14 Equation (1)

3x+y+2z = 11 Equation (2)

2x+3y+z = 11 Equation (3)

Consider the equations (1) and (3)

Equation (1)          x+2y+3z = 14

Equation (3)   *3=6x+9y+3z = 33

Subtracting the equation (1) and equation (3) = -5x -7y = -19

                                                                        = 5x+7y = 19    Equation (4)

Consider the equation (2) and (3)

Equation (1)                3x+y+2z = 11

Equation (3)                 4x+6y+2z = 22

Subtracting equation (1) and equation (3) = -x -5y = -11

                                                                   = x+5y = 11   Equation (5)

Consider the equation (4) and (5)

Equation (4)                5x+7y = 19

Equation (5)         * 3=5x+25y = 55

Subtracting equation (4) and (5)   = -18y = -36

Divide -18 on both sides of the equation

`(-18y)/-18` = `(-36)/-18`

y = 2

Substitute y = 2 in (5) we get

x+5(2)+z = 11

x+10 = 11

Subtract 10 on both sides of the equation

x+10 -10 = 11-10

x = 1

Substitute x = 1, y = 2 in (3) we get

2(1)+3(2)+z = 11

 2+6+z = 11

8+z = 11

Subtract 8 on both sides of the equation

8 – 8 + z = 11 – 8

z = 3

The solution is x = 1, y = 2, z = 3.

Type in algebra 2 questions Practice problems and solutions:

Problem 1: Solve 5x+6 = 3x-6

Solution: -6

Problem 2: Solve 3x – 3y +4z = 14; -9x -6y +2z = 1; 6x+3y+z = 5

Solution: x = 1, y = -1, z = 2.

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