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subject: Basic Algebraic Geometry [print this page]


Basic Algebraic Geometry,basic concepts in algebraic geometry

In geometry, algebra is used to study the numerical numbers and letters. The basic algebra has the polynomial, expressions and equations. The algebraic geometry occupies a central place in modern mathematics and has the multiple conceptual connections with such diverse fields as complex analysis, topology and number theory. Here we will see about the basic algebra with properties and examples. (Source:Wikipedia).

Basic Concepts

The basic concepts in algebraic geometry:

1. Quadratic Equation:

In the algebraic geometry , a quadratic equation in a variable of A. An equation format is as follows,

Ax2+by+c=0

2.Polynomial:

A polynomial of quantity one is known as a linear polynomial.

3.Algebraic Identities:

The form (x+y)2=x2+2xy+y2 is known as the algebraic identities.

4. Algebraic expressions:

It is a separation of polynomial can be representing by a letters and exponent.

Examples

1)6x=36. Find the x value.

Solution:

The given expression is 6x=36

We find the x value only. So we can move the 6 into right side.

So we get x=36/6

X=6

2)Multiply the following equation .(3x+5)(2x2-2x-6)

Solution:

The given equation is (3x+5)(2x2-2x-6).

Now multiply 1st value in the first group with all values of second group. Then multiply the 2nd value in the first group with all values in the second group.

That is 3x(2x2-2x-6)+ 5(2x2-2x-6)

=6x3-6x2-18x+10x2-10x-30

Group the related terms

=6x3+4x2-28x-30

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We can simplify the above equation.We can divide the above equation with the value 2.

So the final equation is 3x3+2x2-14x-15.

3)(8x + 3y "" 8z)

Solution:

Evaluating the given expression (8x + 3y "" 8z) with (x + y + z) ,

From the expression we have,

x = 8x

y = 3y

z= -8z

(8x + 3y "" 8z) = (8x) + (3y) - (8z) + 2(8x) (3y) + 2(3y)(-8z) + 2(-8z)(8x)

So the equation becomes as,

= 8x + 3b + 8z + 48xy -48yz -128za.

These are the examples of algebraic geometry.

Formulas from Algebra and Geometry

triangles

Area = 1 bh

2

law of cosines

c2 = a2 + b2 - 2abcosC

Pythagorean Theorem

a2 + b2 = c2

equilateral triangle

___

A = s2(3)

4

Rectangles

A = bh

parallelograms

A = bh

trapezoids

A = 1 h(b1 + b2)

2

ellipses

A = ab

Circumference = 2 ((a2 + b2)/2)

cones

V = 1r2h

3

circles

Area = r2

Circumference = 2r

Spheres

V = 4r3

3

Surface Area = 4r2

by: Smith




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