Solving Quadratic Formula

Quadratic Equation:

An equation which has one or more terms are squared but no higher power in terms, having the syntax, ax2+bx+c =0 where a represents the numerical coefficient of x2, b represents the numerical coefficient of x, and c represents the constant numerical term.

Types of quadratic equation

Pure quadratic equation:

The numerical coefficient cannot be zero. If b=0 then the quadratic equation is called as a ‘pure’ quadratic equation

Complete quadratic equation:

If the equation having x and x2 terms such an equation is called a ‘complete’ quadratic equation. The constant numerical term ‘c’ may or may not be zero in a complete quadratic equation. Example, x2 + 5x + 6 = 0 and 2x2 - 5x = 0 are complete quadratic equations.


Quadratic Equation Formula


The quadratic equation has the solutions ax2+bx+c =0

x =√(b2-4ac)/2a

Consider the general quadratic equation

ax2+bx+c =0

With a`!=` 0. First divide both sides of the equation by a to get

x2+b/a x + c/a =0

which leads to

x2+ b/a x = - c/a

Next complete the square by adding ((b)/(2a) )2to both sides

X2+ ((b)/(a) )x+((b)/(2a) )2 = -((c)/(a) )+ ((b)/(2a) )2

(x+(b)/(2a) )2=-((c)/(a) ) + ((b^2)/(4a^2) )

(x+(b)/(2a) )2 = (b^2-4ac)/(4a^2)

Finally we take the square root of both sides:

x+(b)/(2a) = +-(sqrt(b^2-4ac))/(2a)

or

x =-(b)/(2a)  +-(sqrt(b^2-4ac))/(2a)

The final form of Quadratic Formula is

x =-b+-sqrt(b^2-4ac)/(2a)

The two roots of the equation is

``          -b-(sqrt(b^2-4ac))/(2a)

-b+(sqrt(b^2-4ac))/(2a)


Example Problem on solving Quadratic formula


Example:

Find the roots of the equation by quadratic formula method, x2-10x+25=0

Solution:

Step 1:  From the equation, a = 1, b = - 10 and c = 25.

Step 2:  To Find X:
plug-in the values in the formula below
x = ``

Step 3:  We get the roots, x = ``
x = 5 and x = 5
which means x1 = 5 and x2 = 5.

Here x = 5 is root of the equation.

learning problems in probability

Introduction:

Learning Probability is also one part of mathematics subject. Probability disturbed with investigation of approximate phenomena. Learning the fundamental things of the probability is random variables, stochastic processes, and events. Mathematical abstraction of the non-deterministic events or calculated quantities that may moreover be single occurrences or evolve over time in an apparently random fashion. While a coin toss or roll of a die is a approximate event.

Let us some example problems in probability with solved solutions.

Learning example problems in probability:

Learning Example problem 1:

Three dice are rolled once. What is the chance that the sum of the expression numbers on the three dice is greater than 15?

Solution:

In probability problems while three dice are rolled,

the trial space S = {(1,1,1), (1,1,2), (1,1,3) ...(6,6,6)}.

S contains 6 × 6 × 6 = 216 outcomes.

Let A be the event of receiving the sum of appearance numbers greater than 15.

A = { (4,6,6), (6,4,6), (6,6,4), (5,5,6), (5,6,5), (6,5,5), (5,6,6), (6,5,6),

(6,6,5), (6,6,6)}.

N (S) = 216 and n (A) = 10.

Therefore P (A) equal to n (A) / n(S)

= 10/ 216

=5 / 108.


Probability Problem 2:


Learning Example problem 2:

In a vehicle Stand there are 100 vehicles, 60 of which are cars, 30 are mini truck and the rest are Lorries. If each vehicle is uniformly likely to leave, find out the probability in this problem.

a) Mini truck departure first.

b) Lorry departure first.

c) Car leaving second if either a lorry or mini truck had gone first.

Solution:

a) Let S be the model space and A be the event of a mini truck leaving first.

n (S) = 100

n (A) = 30

Probability of a mini truck departure first:

P (A) = 30 / 100 = 3 / 10.

b) Let B be the event of a lorry departure first.

n (B) = 100 – 60 – 30 = 10

Probability of a lorry departure first:

P (B) = 10 / 100 = 1 / 10.

c) If moreover a lorry or mini truck had gone first, then there would be 99 vehicles totally left there, 60 of which are cars. Let T be the trial space and C be the occurrence of a car departure.

n(T) = 99

n(C) = 60

Probability of a car departure after a lorry or mini truck has gone:

P(C) = 60 / 99 = 20 / 33.

Learning Probability Density Functions

The Learning Probability density functions are known as Gaussian functions. A probability density function of a continues random variable is defined as a function that describes the virtual probability for that random variable to occur at a given point within the using space. The Learning of probability density functions is same for the probability distribution functions. In this article we see the learning  the definition of probability density functions and some example problem for the probability density functions.

Definition of Learning the Probability Density Functions:


Probability density function is defined as continues random variable functions f(x). this functions satisfies the following properties.

Probability functions limits between a and b

P(a≤x≤b) = ∫ a to b  f(x) . dx

A probability density function has only real for the real value.

F(x) ≥ 0

Integral of the probability density functions is 1.

∫ -oo to oo f(x) dx  = 1

Learning the fundamental properties of probability density functions:

F(x) is continues random functions

P(a<= x <b) =  P(a < x<b) = P( a< x <= b)

= P(a<=X<b) =

Where f is probability density distribution functions

In the differentional functional of the probability density functions, we have

P(x<X<=x+dx) = F(x+dx)-F(x) = dF(x)=F'(x)dx = f(x)dx

Where

X is Known as probability differential functions.


Learning the some example problems for the probability density functions:


Example 1:

If the probability density function of a random variable is given by f(x) = H (x – x^3);  0 < x <1 .find the H.

Solution:

Since ,

F(x) is the probability density functions means,

F(x)= ∫ a to b f(x) dx

Substitute the values,

H [ x^2/2 – x^4/4] limits 0 to1 = 1

H [ 1/2- 1/4] = 1

1 After simplify, We get

H= 4

Example-2

If the probability density function of a random variable is given by f(x) = H (1– x^5);  0 < x <1 .find the H.

Solution:

Since ,

F(x) is the probability density functions means,

F(x)= ∫ a to b f(x) dx

Substitute the values,

H [ x – x^6/6] limits 0 to1 = 1

H [ 1- 1/6] = 1

1 After simplify, We get

H= 6/5 = 1.2

Practice problem of  learning the probability density functions:

Problem -1:

A continuous random variable X follows the probability law, f(x) =H x (x – x^2 ), 0 < x < 1and 0 for elsewhere. Find k

Answer: H = 12

Problem -2:

A continuous random variable X follows the probability law, f(x) =H x^2 (x– x^4 ), 0 < x < 1and 0 for elsewhere. Find k

Answer: H = 9.33

Learn Operations on Functions

Generally a function is represented in terms of  f(x).

Functions behave exactly as one would expect with regard to the four basic operations of algebra (addition, subtraction, multiplication, and division). When functions are combined by these operations, though, the domain of the new combined function is only the elements that were shared by the domains of the original functions.

Operations on functions :

As we add, subtract, multiply the numbers, we can add , subtract, multiply, divide the functions. There is one operation more called the composition of functions.

Let us learn through examples, that would be easier to understand.

Consider the two functions,

f(x) = x +2

g (x) = x2 -5


Operations on functions : Addition


f(x) = x +2

g (x) = x^2 -5

(f+g) (x) = x +2 +x2 -5

= x2+x -5 +2

= x2 + x -3

(f+g) (2) = 22 +2 -3 = 6 -3 =3


Operations on functions : Subtraction


f(x) = x +2

g (x) = x^2 -5

(f-g) (x) = x+2 – (x2 -5)

= x2+x +2 -5

=x2 +x -3

(f-g)(2) = 2 +2 - 22 -5 = -5


Operations on functions : Multiplication


f(x) = x +2

g (x) = x^2 -5

(f)(g)(x) = (x+2) (x2-5)

= x3 -5x +2x2 -10

= x3 +2x2 -5x -10

(f)(g)(2) = (2+2) (22-5) = -4


Operations on functions : Division


f(x) = x +2

g (x) = x2 -5

g(x) /f(x) =

=  [x2 -5] / [x+2]

g (2) / f (2 ) = (22 - 5)/ (2+2)

=(4 -5)/ (2+2) = -1 / 4


Operations on functions : Composition of Function


f(x) = x +2

g (x) = x^2 -5

(f o g)(x) read as f of g of x, we can say  that first, we find g(x) and then we find f(x) for the result that we got from g(x)

The concept is simple. First, the value of g at x is taken, and then the value of f at that value is taken

g(x) = x2 -5

(f o g)(x)= f (x2-5) = x2 -5 +2 = x2 -3

(f o g)(2) =

= g(2) =22 – 5 = -1

= f(-1) = -1+2 =1

Trigonometric Functions Unit Circle

Unit circle:

In mathematics, a unit circle is a circle with a radius of one. Frequently, except in trigonometry, the unit circle is the circle of radius one centered at the origin (0, 0) in the Cartesian coordinate system in the Euclidean plane. The unit circle is denoted as s1; the generalization to higher dimensions is the unit sphere.

Trigonometric functions on the unit circle:

All of the trigonometric functions of the angle θ can be modified by geometrically in terms of a unit circle centered at O.

The trigonometric functions of cosine and sine can be defined on the unit circle as follows. If (x, y) is a point of the unit circle, and if the ray from the origin (0, 0) to (x, y) makes an angle t from the positive x-axis, (where counterclockwise turning is positive), then

cos(t) = x

sin(t) = y

The equation x2 + y2 = 1 gives the relation

cos2(t) + sin2(t) = 1

The unit circle also operated that sine and cosine are periodic functions, with the identities

cos t = cos(2πk + t)

sin t = sin(2πk + t)

For any integer k

Area & Circumference:

1. Length of circumference:

The circumference’s length is related to the radius (r) by

c = 2*π*r

For unit circle circumference is 2*(because r = 1).

2. Area enclosed:

Area of the circle = π × area of the shaded square

Main article: Area of a disk

The area of the circle is π multiplied with the radius squared:

A = π *r2

For unit circle area is


Application:


Circle group:

The complex numbers can be pointed with points in the Euclidean plane, namely the number a + bi is identified with the point (a, b). Under this identification, the unit circle is a group under multiplication, called the circle group. It has main applications in mathematics and science.

Application of the Unit Circle:

• It is used to understand the relationship between the unit circle and the trigonometric (sine and cosine) functions.

• Because of this reason it is used to solve a real-world application.

Multiplication Solver

Multiplication (symbol "×") solver is the mathematical operation of scaling one number by another. It is one of the four basic operations in a elementary arithmetic (the others being addition, subtraction and division) (Source.Wikipedia). Multiplication solver  there are two jobs such that one of them can be completed in m ways, and when it has been completed in any one of these m ways, second job can be completed in n ways; then the two jobs in succession can be completed in m × n ways.

Examples for multiplication solver:


Rules for multiplication solver:

•    The multiplication solver of two positive integers is a positive integer.

•    The multiplication solver of a positive integer and the negative integer is a negative integer.

•    The multiplication solver of two negative integers is a positive integer.

Example 1:

Multiply 283 by 104

Solution:

283 × 104 means 104 times 283

or 100 times 283 + four time 283

or 28300 + [4* 283]

or 29432

∴ 283 × 104 = 29432 [multiplication rules: The product of two positive integers is a positive integer.]

Example 2:

Multiply 750 × 98

Solution:

750 × 98 means 98 times 750

or 100 times 750 – two time 750

or 75000 – 1500

or 73500

∴ 750 × 98 = 73500      [multiplication rules: The product of a positive integer and the negative integer is a negative integer.]

Example 3:

Find the value of 10 × 5 + 12 × 5 + 14 × 5

Solution:

10 × 5 + 12 × 5 + 14 × 5 means

10 times 5 + 12 times 5 + 14 times 5

or 36 times 5

or 36 × 5

or 180

∴ The value = 180


Understanding Example 4:

Multiply 843 by 55 by ordinary method

Solution:

8 4 3 ×

5 5
..............
4215
...................
4 6 3 6 5              [multiplication rules: The product of two positive integers is a positive integer.]
.......................

Example 5:

Find the product of (a) (+ 6) × (– 6) (b) (– 13) × (+ 13) (c) (– 15) × (– 4)

Solution:

(a) (+ 6) × (– 6) = – 36 (positive × negative = negative)

(b) (– 13) × (+ 13) = – 169 (negative × positive = negatives)

(c) (– 15) × (– 5) = + 75 (negative × negative = positive)

Example 5:

Find the product of (a) (– 4) × (+ 5) × (+ 2) (b) (– 2) × (– 5) × (+ 10) (c) (– 4) × (– 2) × (– 6)

Solution:

(a) (– 4) × (+ 5) × (+ 2) (b) (– 2) × (– 5) × (+ 10) (c) (– 4) × (– 2) × (– 6)

= (– 20) × (+ 2) = (+ 10) × (+ 10) = (+ 8) × (– 6) [The product of a positive integer and the negative integer is a negative integer. multiplication rules]

= – 40 = + 100 = – 48

Example 6 :

Multiply  2 and 3

Solution:

By repeated addition we get the following:

2 × 4 = 2 + 2 + 2+2 = 8.

Practice problem for multiplication solver:


1.Multiply  2 and 5

Answer:10

2.Multiply  100 and 108

Answer:10800

learning area theorems

There are so many theorems on areas and deductions from that main theorem.  Any planar closed figure has an area.  If the figure is a noted one like triangle, rectangle, circle, polygon etc. we have specific formula so that we can find out areas by applying formula.

For Geometric closed figures like quadrilateral, irregular polygons we use intersecting lines to divide into triangles and add all areas of the triangle, so that all the triangles taken for this purpose are different segments of the polygon.

In this article, we study about how to calculate area for regular polygons.

Area of a polygon:

A polygon is a closed curve consisting of line segments in a plan which are called sides of the polygon.  There are no restrictions to the number of sides of a polygon in a plane.  Starting from 3, it can have any no of sides up to infinity.

Regular polygons are polygons which have equal sides and equal angles.  Regular polygons are special types of polygons with certain extra properties.

For an irregular polygon, no standard formula but can be considered as the sum of all triangles formed by joining the centroid of the polygon to each vertex.  If there are n sides, n triangles will be formed.  The full area of the irregular polygon will be the sum of area of all triangles.

Corollary to the above for a regular polygon:

For a regular polygon of n sides we know there exists a centre and from centre lines are drawn to each vertex, the polygon will be a combination of n congruent triangles.  In other words, for a regular polygon the lines drawn from centre divides the polyton into n number of congruent triangles.  If we calculate one triangle area, then total area is calculated by multiplying by number of sides.

So Area of the polygon = n(area of each triangle) = n(1/2*side*height from centre of polygon to any side)

= 1/2 (n*side)*apothem    (Since apothem is the altitude of the side from the centre)

= 1/2 * Perimeter*apothem  (since n multiplied by no of sides give perimeter)


Learning problems solved for regular polygons - area theorem


Learning area theorems:

Problem1:  Find area of a regular hexagon with side 6.

Solution:

Area of hexagon of 6 equal sides = 1/2 (perimeter)(apotnem)

We know that a hexagon when divided into 6 triangles, each triangle is equilateral as central angle is 60.

Area of an equilateral triangle =√ 3 (a^2)/4 = 1.732*6*6/4 = 15.588

Hence area of hexagon = 6*15.588 = 93.528


Applications of regular area theorems in squares and triangles:

This theorem applies for equilateral triangle and square also.

For triangle, centroid is the centre and we know the apothem for the triangle formed with two vertices and one side is 1/3 h where h is the height of the triangle.  So area of the triangle = 3*a*1/2*1/3h = a2h/2 = √3/2a/2*a = √3a^2/4 = Area of the equilateral triangle

For square, the intersection of diagonals is the centroid and the apothem for each triangle is a/2.

So area of square by applying this theorem = 4a*a/2*1/2 =a^2

Problem 2:   Find area of a regular pentagon with side 7.

Solution:  Here we know each angle for a regular pentagon = 540/5 =108.

Hence line joining each vertex to the centre forms an angle of 108/2 = 54 degree with the side.

So 5 triangles are formed with base as 7 and angles two equalling 54 degrees.

Now we can find apothem using this.  Apothem / 1/2 side = tan 54:   or apothem = 7tan 54/2 =7(1.3763)/2

= 4.8173

Area of pentagon = 5*7*4.8173/2 = 168.61/2 =84.303

Thus given side and no of sides, the area of a regular polygon we can find out by using area theorems for polygons.

CONCLUSION:

In this article, we learnt about the area theorems for polygons and worked out problems on that. An interesting thing in a regular polygon is if we know the side we can find out the area of the polygon.