Saturday, December 15, 2018

Statistics

Statistics



Find the mean number of plants per house. Statistics deals with collection, presentation, analysis and interpretation of numerical data.

Arranging data in a order to study their salient features is called presentation of data.

Range of the data is the difference between the maximum and the minimum values of the observations

Table that shows the frequency of different values in the given data is called a frequency distribution table

A frequency distribution table that shows the frequency of each individual value in the given data is called an ungrouped frequency distribution table.

A table that shows the frequency of groups of values in the given data is called a grouped frequency distribution table

The groupings used to group the values in given data are called classes or class-intervals. The number of values that each class contains is called the class size or class width. The lower value in a class is called the lower class limit. The higher value in a class is called the upper class limit.

Class mark of a class is the mid value of the two limits of that class.

Mean of Grouped Data
The mean (or average) of observations, as we know, is the sum of the values of all the observations divided by the total number of observations

Mean x̄ 
=  (f1x1 + f1x1 + L + fnxn
  f1 + f1 + L + fn 

Mean x̄ 
= ∑ fixi/fi
where i varies from 1 to n

We can form ungrouped data into grouped data by forming class-intervals of some width.

It is assumed that the frequency of each classinterval is centred around its mid-point.

Class mark 
= (Upper class limit + Lower class limit)/2

Direct method 
Mean x̄ 
= ∑ fixi/fi

Assumed Mean method 
Mean x̄ 
= a + [∑ fidi]/fi

We can only locate a class with the maximum frequency, called the modal class. The mode is a value inside the modal class, and is given by the formula:
Mode
= l + [(f1 - f0)/((2f1) - f0 - f2)]*h

where l = lower limit of the modal class,
h = size of the class interval (assuming all class sizes to be equal),
f1 = frequency of the modal class,
f0 = frequency of the class preceding the modal class,
f2 = frequency of the class succeeding the modal class.

for finding the median of ungrouped data, we first arrange the data values of the observations in ascending order. Then, if n is odd, the median is the (n+1)/2 th observation. And, if n is even, then the median will be the average of the n/2 th and (n/2)+1 th observations

Less than cumulative frequency distribution:
It is obtained by adding successively the frequencies of all the previous classes including the class against which it is written. The cumulate is started from the lowest to the highest size.

More than cumulative frequency distribution:
It is obtained by finding the cumulate total of frequencies starting from the highest to the lowest class.

To find median class, we find the cumulative frequencies of all the classes and n/2. We now locate the class whose cumulative frequency is greater than (and nearest to) n/2.

After finding the median class, we use the following formula for calculating the median.

Median 
= l + [((n/2) - cf)/f]*h

where 
l = lower limit of median class,
n = number of observations,
cf = cumulative frequency of class preceding the median class,
f = frequency of median class,
h = class size (assuming class size to be equal).

There is a empirical relationship between the three measures of central tendency :
3 Median = Mode + 2 Mean

Representing a cumulative frequency distribution graphically as a cumulative frequency curve, or an ogive of the less than type and of the more than type.

The median of grouped data can be obtained graphically as the x-coordinate of the point of intersection of the two ogives for this data.

Statistics



Find the mean number of plants per house. Statistics deals with collection, presentation, analysis and interpretation of numerical data.

Arranging data in a order to study their salient features is called presentation of data.

Range of the data is the difference between the maximum and the minimum values of the observations

Table that shows the frequency of different values in the given data is called a frequency distribution table

A frequency distribution table that shows the frequency of each individual value in the given data is called an ungrouped frequency distribution table.

A table that shows the frequency of groups of values in the given data is called a grouped frequency distribution table

The groupings used to group the values in given data are called classes or class-intervals. The number of values that each class contains is called the class size or class width. The lower value in a class is called the lower class limit. The higher value in a class is called the upper class limit.

Class mark of a class is the mid value of the two limits of that class.

Mean of Grouped Data
The mean (or average) of observations, as we know, is the sum of the values of all the observations divided by the total number of observations

Mean x̄ 
=  (f1x1 + f1x1 + L + fnxn
  f1 + f1 + L + fn 

Mean x̄ 
= ∑ fixi/fi
where i varies from 1 to n

We can form ungrouped data into grouped data by forming class-intervals of some width.

It is assumed that the frequency of each classinterval is centred around its mid-point.

Class mark 
= (Upper class limit + Lower class limit)/2

Direct method 
Mean x̄ 
= ∑ fixi/fi

Assumed Mean method 
Mean x̄ 
= a + [∑ fidi]/fi

We can only locate a class with the maximum frequency, called the modal class. The mode is a value inside the modal class, and is given by the formula:
Mode
= l + [(f1 - f0)/((2f1) - f0 - f2)]*h

where l = lower limit of the modal class,
h = size of the class interval (assuming all class sizes to be equal),
f1 = frequency of the modal class,
f0 = frequency of the class preceding the modal class,
f2 = frequency of the class succeeding the modal class.

for finding the median of ungrouped data, we first arrange the data values of the observations in ascending order. Then, if n is odd, the median is the (n+1)/2 th observation. And, if n is even, then the median will be the average of the n/2 th and (n/2)+1 th observations

Less than cumulative frequency distribution:
It is obtained by adding successively the frequencies of all the previous classes including the class against which it is written. The cumulate is started from the lowest to the highest size.

More than cumulative frequency distribution:
It is obtained by finding the cumulate total of frequencies starting from the highest to the lowest class.

To find median class, we find the cumulative frequencies of all the classes and n/2. We now locate the class whose cumulative frequency is greater than (and nearest to) n/2.

After finding the median class, we use the following formula for calculating the median.

Median 
= l + [((n/2) - cf)/f]*h

where 
l = lower limit of median class,
n = number of observations,
cf = cumulative frequency of class preceding the median class,
f = frequency of median class,
h = class size (assuming class size to be equal).

There is a empirical relationship between the three measures of central tendency :
3 Median = Mode + 2 Mean

Representing a cumulative frequency distribution graphically as a cumulative frequency curve, or an ogive of the less than type and of the more than type.

The median of grouped data can be obtained graphically as the x-coordinate of the point of intersection of the two ogives for this data.

Triangles

Triangles



All congruent figures are similar, but it does not mean that all similar figures are congruent.

Two polygons of the same number of sides are similar, if:
Their corresponding angles are equal.
Their corresponding sides are in the same ratio.

Two triangles are similar, if:
Their corresponding angles are equal.
Their corresponding sides are in the same ratio.

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar.

If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

If in two triangles, sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar.

If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.

The ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

Triangles



All congruent figures are similar, but it does not mean that all similar figures are congruent.

Two polygons of the same number of sides are similar, if:
Their corresponding angles are equal.
Their corresponding sides are in the same ratio.

Two triangles are similar, if:
Their corresponding angles are equal.
Their corresponding sides are in the same ratio.

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar.

If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

If in two triangles, sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar.

If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.

The ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

Arithmetic Progressions

Arithmetic Progressions



(i) 1, 2, 3, 4 ...
(ii) - 5, - 1, 3, 7 ...
(iii) 1/3, 5/3, 9/3, 13/3 ....
(iv) 0.6, 1.7, 2.8, 3.9 ...
An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant

The common difference of the AP is a sequence is always fixed. It can be positive, negative or zero.

Let us denote the first term of an AP by a1, second term by a2, . . ., nth term by anand the common difference by d.

If the AP series has last term then it s finite Arithmetic Progression and if the AP series has infinite then it is called the Inifinite Arithmetic Progression

General form of AP is
a, a+d, a+2d, a+3d, a+4d, .... 

nth term an of the AP with first term a and common difference d is given by 
an = a + (n – 1) d
an is also called the general term of the AP.

Sum of n term of Arithmetic progression
Sn = (n/2)[2a + (n - 1)d]

If l is the last term of the finite AP, say the nth term, then the sum of all terms of the AP is given
S = (n/2)[a + l]

Arithmetic Progressions



(i) 1, 2, 3, 4 ...
(ii) - 5, - 1, 3, 7 ...
(iii) 1/3, 5/3, 9/3, 13/3 ....
(iv) 0.6, 1.7, 2.8, 3.9 ...
An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant

The common difference of the AP is a sequence is always fixed. It can be positive, negative or zero.

Let us denote the first term of an AP by a1, second term by a2, . . ., nth term by anand the common difference by d.

If the AP series has last term then it s finite Arithmetic Progression and if the AP series has infinite then it is called the Inifinite Arithmetic Progression

General form of AP is
a, a+d, a+2d, a+3d, a+4d, .... 

nth term an of the AP with first term a and common difference d is given by 
an = a + (n – 1) d
an is also called the general term of the AP.

Sum of n term of Arithmetic progression
Sn = (n/2)[2a + (n - 1)d]

If l is the last term of the finite AP, say the nth term, then the sum of all terms of the AP is given
S = (n/2)[a + l]

Pair of Linear Equations in two variables

Pair of Linear Equations in two variables



An equation which can be put in the form ax + by + c = 0, where a, b and c are real numbers, and a and b are not both zero, is called a linear equation in two variables x and y. (We often denote the condition a and b are not both zero by a2+ b2 ≠ 0)

For example, let us substitute x = 2 and y = 2 in the left hand side (LHS) of the equation 3x + 4y = 14
Then
LHS = 3(2) + 4(2) 
= 6 + 8 
= 14,
which is equal to the right hand side (RHS) of the equation.
∴ x = 2 and y = 2 is a solution of the equation 3x + 4y = 14

Now let us substitute x = 1 and y = 1 in the equation 3x + 4y = 14. 
Then,
LHS = 3(1) + 4(1) 
= 3 + 4 
= 7
which is not equal to the RHS
∴ x = 1 and y = 1 is not a solution of the equation.

Geometrically, what does this mean? It means that the point (2, 2) lies on the line representing the equation 3x + 4y = 14, and the point (1, 1) does not lie on it. So, every solution of the equation is a point on the line representing it.

In fact, this is true for any linear equation, that is, each solution (x, y) of a linear equation in two variables, ax + by + c = 0, corresponds to a point on the line representing the equation, and vice versa.

The general form for a pair of linear equations in two variables x and y is
a1x + b1y + c1 = 0 and
a2x + b2y + c2 = 0,

where a1, b1, c1, a2, b2, c2 are all real numbers and
a12 + b12 ≠ 0, 
a22 + b22 ≠ 0.

Example 5x + 34y – 7 = 0 and 5x – 22y + 9 = 0

To summarise the behaviour of lines representing a pair of linear equations in two variables and the existence of solutions
(i) x – y = 1 and x + y = 3 (The lines intersect) 

(ii) the lines may be parallel. In this case, the equations have no solution (inconsistent pair of equations).
x + y = 2 and x + y = 3 

(iii) the lines may be coincident. In this case, the equations have infinitely many solutions [dependent (consistent) pair of equations].
x + y = 2 and 2x + 2y = 4 

Lines represented by
a1x + b1y + c1 = 0 and
a2x + b2y + c2 = 0,

(i) intersecting, then
a1/a2 ≠ b1/b2

(ii) coincident, then
a1/a2 = b1/b2 = c1/c2

(iii) parallel, then
a1/a2 = b1/b2 ≠ c1/c2

Substitution Method : We shall explain the method of substitution by taking some examples.
Solve the following pair of equations by substitution method:
x – y = 1 ⇒ (1)
x + y = 3 ⇒(2)
Solution :
Step 1 : 
We pick either of the equations and write one variable in terms of the other.
Let us consider the Equation (2)
x + y = 3
and write it as x = 3 – y
Step 2 : 
Substitute the value of x in Equation (1)
We get
(3 – y) – y = 1
3 – 2y = 1
2y = 2
∴ y = 1
Step 3 : 
Substituting this value of y in Equation (1), we get
x = 2

Real Life application:

You have a specific area of plot where you are building your home. At one corner you plan to have experimental room. You don't know the dimension of it, but there is a relation where in area of a room gets reduced by 9 square units when length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the experimental room
Elimination Method
2x – y = 3 ⇒ (1)
x + y = 3 ⇒(2)
Step 1 : 
Multiply Equation (1) by 1 and Equation (2) by 2 to make the coefficients of x equal. Then we get the equations:
2x – y = 3 ⇒(3)
2x + 2y = 6 ⇒(4)
Step 2 : 
Subtract Equation (3) from Equation (4) to eliminate x, because the coefficients of y are the same. So, we get
3y = 3
y = 1
Step 3 : 
Substituting this value of y in (1), we get
x = 2
Cross multiplication
Lines represented by
a1x + b1y + c1 = 0 and
a2x + b2y + c2 = 0

We can express it as
x/(b1c2 - b2c1)
= y/(c1a2 - c2a1)
= 1/(a1b2 - a2b1

Real Life application:

Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?

Let the speed of 1st car and 2nd car be u and v
Respective speed of both cars while they are travelling in same direction = (u - v) km/h
Respective speed of both cars while they are travelling towards each other = (u + v) km/h
Given: When the travel in same direction they meet after 5 hours

but require 1 hour when they travel towards each other.

Place is 100km apart
5(u - v) = 100
u - v = 20 km/h ⇒ (i)
1(u + v) = 100 ⇒ (ii)
Adding both the equations, we get
2u = 120 
u = 60 km/h ⇒ (iii)
Substituting this value in equation (ii), we obtain
v = 40 km/h
∴ speed of one car = 60 km/h and speed of other car = 40 km/h

Pair of Linear Equations in two variables



An equation which can be put in the form ax + by + c = 0, where a, b and c are real numbers, and a and b are not both zero, is called a linear equation in two variables x and y. (We often denote the condition a and b are not both zero by a2+ b2 ≠ 0)

For example, let us substitute x = 2 and y = 2 in the left hand side (LHS) of the equation 3x + 4y = 14
Then
LHS = 3(2) + 4(2) 
= 6 + 8 
= 14,
which is equal to the right hand side (RHS) of the equation.
∴ x = 2 and y = 2 is a solution of the equation 3x + 4y = 14

Now let us substitute x = 1 and y = 1 in the equation 3x + 4y = 14. 
Then,
LHS = 3(1) + 4(1) 
= 3 + 4 
= 7
which is not equal to the RHS
∴ x = 1 and y = 1 is not a solution of the equation.

Geometrically, what does this mean? It means that the point (2, 2) lies on the line representing the equation 3x + 4y = 14, and the point (1, 1) does not lie on it. So, every solution of the equation is a point on the line representing it.

In fact, this is true for any linear equation, that is, each solution (x, y) of a linear equation in two variables, ax + by + c = 0, corresponds to a point on the line representing the equation, and vice versa.

The general form for a pair of linear equations in two variables x and y is
a1x + b1y + c1 = 0 and
a2x + b2y + c2 = 0,

where a1, b1, c1, a2, b2, c2 are all real numbers and
a12 + b12 ≠ 0, 
a22 + b22 ≠ 0.

Example 5x + 34y – 7 = 0 and 5x – 22y + 9 = 0

To summarise the behaviour of lines representing a pair of linear equations in two variables and the existence of solutions
(i) x – y = 1 and x + y = 3 (The lines intersect) 

(ii) the lines may be parallel. In this case, the equations have no solution (inconsistent pair of equations).
x + y = 2 and x + y = 3 

(iii) the lines may be coincident. In this case, the equations have infinitely many solutions [dependent (consistent) pair of equations].
x + y = 2 and 2x + 2y = 4 

Lines represented by
a1x + b1y + c1 = 0 and
a2x + b2y + c2 = 0,

(i) intersecting, then
a1/a2 ≠ b1/b2

(ii) coincident, then
a1/a2 = b1/b2 = c1/c2

(iii) parallel, then
a1/a2 = b1/b2 ≠ c1/c2

Substitution Method : We shall explain the method of substitution by taking some examples.
Solve the following pair of equations by substitution method:
x – y = 1 ⇒ (1)
x + y = 3 ⇒(2)
Solution :
Step 1 : 
We pick either of the equations and write one variable in terms of the other.
Let us consider the Equation (2)
x + y = 3
and write it as x = 3 – y
Step 2 : 
Substitute the value of x in Equation (1)
We get
(3 – y) – y = 1
3 – 2y = 1
2y = 2
∴ y = 1
Step 3 : 
Substituting this value of y in Equation (1), we get
x = 2

Real Life application:

You have a specific area of plot where you are building your home. At one corner you plan to have experimental room. You don't know the dimension of it, but there is a relation where in area of a room gets reduced by 9 square units when length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the experimental room
Elimination Method
2x – y = 3 ⇒ (1)
x + y = 3 ⇒(2)
Step 1 : 
Multiply Equation (1) by 1 and Equation (2) by 2 to make the coefficients of x equal. Then we get the equations:
2x – y = 3 ⇒(3)
2x + 2y = 6 ⇒(4)
Step 2 : 
Subtract Equation (3) from Equation (4) to eliminate x, because the coefficients of y are the same. So, we get
3y = 3
y = 1
Step 3 : 
Substituting this value of y in (1), we get
x = 2
Cross multiplication
Lines represented by
a1x + b1y + c1 = 0 and
a2x + b2y + c2 = 0

We can express it as
x/(b1c2 - b2c1)
= y/(c1a2 - c2a1)
= 1/(a1b2 - a2b1

Real Life application:

Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?

Let the speed of 1st car and 2nd car be u and v
Respective speed of both cars while they are travelling in same direction = (u - v) km/h
Respective speed of both cars while they are travelling towards each other = (u + v) km/h
Given: When the travel in same direction they meet after 5 hours

but require 1 hour when they travel towards each other.

Place is 100km apart
5(u - v) = 100
u - v = 20 km/h ⇒ (i)
1(u + v) = 100 ⇒ (ii)
Adding both the equations, we get
2u = 120 
u = 60 km/h ⇒ (iii)
Substituting this value in equation (ii), we obtain
v = 40 km/h
∴ speed of one car = 60 km/h and speed of other car = 40 km/h

Polynomial

Degree of the polynomial
The highest power of x in p(x) is called the degree of the polynomial.

Polynomial of degree 1 is known as linear polynomial.
Standard form is ax+b, where a and b are real numbers and a ≠ 0.
4x+1 is a linear polynomial. 

Polynomial of degree 2 is known as quadratic polynomial.
Standard form is ax2 + bx + c, where a, b and c are real numbers and a ≠ 0
x2 + 4x + 6 is an example for quadratic polynomial.

Polynomial of degree 3 is known as cubic polynomial.
Standard form is ax3 + bx2 + cx + d, where a, b, c and d are real numbers and a ≠ 0.
x3 + 2x + 6 is an example for cubic polynomial.

If p(x) is a polynomial in x, and if k is any real number, then the value obtained by
replacing x by k in p(x), is called the value of p(x) at x = k, and is denoted by p(k).

p(x) = x3 + 4x + 5
at x = -1 we have
p(-1) = (-1)3 + 4(-1) + 5
= -1 - 4 + 5
= 0
As p(–1) = 0, –1 is called the zeroes of the quadratic polynomial x3 + 4x + 5. More generally, a real number k is said to be a zero of a polynomial p(x), if p(k) = 0 the zeroes of a quadratic polynomial ax2 + bx + c, a ≠ 0, are precisely the x-coordinates of the points where the parabola representing y = ax2 + bx + c intersects the x-axis.

In general, given a polynomial p(x) of degree n, the graph of y = p(x) intersects the x- axis at atmost n points. Therefore, a polynomial p(x) of degree n has at most n zeroes.

If α and β are zeros of the polynomial then
Sum of its zeroes α + β
= -(Coefficient of x)/(Coefficient of x2) = -b/a

Product of zeroes = αβ
Constant term/Coefficient of x2 = c/a

If α, β, γ are the zeroes of the cubic polynomial ax3 + bx2 + cx + d, then

α + β + γ = -b/a
αβ + αγ + βγ = c/a
αβγ = -d/a

Division Algorithm
If p(x) and g(x) are any two polynomials with g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that
p(x) = g(x) * q(x) + r(x)
where r(x) = 0 or degree of r(x) < degree of g(x).

Degree of the polynomial
The highest power of x in p(x) is called the degree of the polynomial.

Polynomial of degree 1 is known as linear polynomial.
Standard form is ax+b, where a and b are real numbers and a ≠ 0.
4x+1 is a linear polynomial. 

Polynomial of degree 2 is known as quadratic polynomial.
Standard form is ax2 + bx + c, where a, b and c are real numbers and a ≠ 0
x2 + 4x + 6 is an example for quadratic polynomial.

Polynomial of degree 3 is known as cubic polynomial.
Standard form is ax3 + bx2 + cx + d, where a, b, c and d are real numbers and a ≠ 0.
x3 + 2x + 6 is an example for cubic polynomial.

If p(x) is a polynomial in x, and if k is any real number, then the value obtained by
replacing x by k in p(x), is called the value of p(x) at x = k, and is denoted by p(k).

p(x) = x3 + 4x + 5
at x = -1 we have
p(-1) = (-1)3 + 4(-1) + 5
= -1 - 4 + 5
= 0
As p(–1) = 0, –1 is called the zeroes of the quadratic polynomial x3 + 4x + 5. More generally, a real number k is said to be a zero of a polynomial p(x), if p(k) = 0 the zeroes of a quadratic polynomial ax2 + bx + c, a ≠ 0, are precisely the x-coordinates of the points where the parabola representing y = ax2 + bx + c intersects the x-axis.

In general, given a polynomial p(x) of degree n, the graph of y = p(x) intersects the x- axis at atmost n points. Therefore, a polynomial p(x) of degree n has at most n zeroes.

If α and β are zeros of the polynomial then
Sum of its zeroes α + β
= -(Coefficient of x)/(Coefficient of x2) = -b/a

Product of zeroes = αβ
Constant term/Coefficient of x2 = c/a

If α, β, γ are the zeroes of the cubic polynomial ax3 + bx2 + cx + d, then

α + β + γ = -b/a
αβ + αγ + βγ = c/a
αβγ = -d/a

Division Algorithm
If p(x) and g(x) are any two polynomials with g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that
p(x) = g(x) * q(x) + r(x)
where r(x) = 0 or degree of r(x) < degree of g(x).

Real Numbers:

Real Numbers:
Real numbers can be both positive or negative, and they are denoted by the symbol R. Numbers like a natural number, decimals, and fraction comes under the real number.

Euclids division algorithm is a technique to compute the Highest Common Factor (HCF) of two given positive integers.

To obtain the HCF of two positive integers, say c and d, with c > d, follow the steps below:
Step 1 : Apply Euclid’s division lemma, to c and d. So, we find whole numbers, q and r such that c = dq + r, 0 ≤ r < d.
Step 2 : If r = 0, d is the HCF of c and d. If r ≠ 0, apply the division lemma to d and r.
Step 3 : Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.

Let us see how the algorithm works, through an example first. Suppose we need to find the HCF of the integers 120 and 50. We start with the larger integer, that is, 120. 

Then we use Euclids lemma to get
120 = 50 x 2 + 20
Now consider the divisor 50 and the remainder 20, and apply the division lemma to get
50 = 20 x 2 + 10
Now consider the divisor 20 and the remainder 10, and apply the division lemma to get
20 = 10 x 2 + 0
Notice that the remainder has become zero, and we cannot proceed any further.
We claim that the HCF of 120 and 50 is the divisor at this stage, i.e., 10. You can easily verify this by listing all the factors of 120 and 50. 

Euclids division lemma/algorithm has several applications

A fruitstall has 120 apple and 50 mangoes. He wants to stack them in such a way that each stack has the same number, and they take up the least area of the tray. What is the number of mangoes and apple that can be placed in each stack for this purpose?
Solution : This can be done by trial and error. But to do it systematically, we find HCF (120, 50). Then this number will give the maximum number of fruits in each stack and the number of stacks will then be the least. The area of the tray that is used up will be the least.
Now, let us use Euclids algorithm to find their HCF. We have :
120 = 50 x 2 + 20
50 = 20 x 2 + 10
20 = 10 x 2 + 0
So, the HCF of 120 and 50 is 10.
∴ The fruitseller can make stacks of 10 mangoes and 10 apples

Fundamental Theorem of Arithmetic : 
Every composite number can be expressed ( factorised ) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.

A number s is called irrational if it cannot be written in the form p/q where p and q are integers and q ≠ 0

Rational Numbers and Their Decimal Expansions
Rational numbers have either a terminating decimal expansion or a non-terminating repeating decimal expansion.

Real number is a rational number of the form p/q where the prime factorisation of q is of the form 2n5m and n, m are some non-negative integers.

Real Numbers:
Real numbers can be both positive or negative, and they are denoted by the symbol R. Numbers like a natural number, decimals, and fraction comes under the real number.

Euclids division algorithm is a technique to compute the Highest Common Factor (HCF) of two given positive integers.

To obtain the HCF of two positive integers, say c and d, with c > d, follow the steps below:
Step 1 : Apply Euclid’s division lemma, to c and d. So, we find whole numbers, q and r such that c = dq + r, 0 ≤ r < d.
Step 2 : If r = 0, d is the HCF of c and d. If r ≠ 0, apply the division lemma to d and r.
Step 3 : Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.

Let us see how the algorithm works, through an example first. Suppose we need to find the HCF of the integers 120 and 50. We start with the larger integer, that is, 120. 

Then we use Euclids lemma to get
120 = 50 x 2 + 20
Now consider the divisor 50 and the remainder 20, and apply the division lemma to get
50 = 20 x 2 + 10
Now consider the divisor 20 and the remainder 10, and apply the division lemma to get
20 = 10 x 2 + 0
Notice that the remainder has become zero, and we cannot proceed any further.
We claim that the HCF of 120 and 50 is the divisor at this stage, i.e., 10. You can easily verify this by listing all the factors of 120 and 50. 

Euclids division lemma/algorithm has several applications

A fruitstall has 120 apple and 50 mangoes. He wants to stack them in such a way that each stack has the same number, and they take up the least area of the tray. What is the number of mangoes and apple that can be placed in each stack for this purpose?
Solution : This can be done by trial and error. But to do it systematically, we find HCF (120, 50). Then this number will give the maximum number of fruits in each stack and the number of stacks will then be the least. The area of the tray that is used up will be the least.
Now, let us use Euclids algorithm to find their HCF. We have :
120 = 50 x 2 + 20
50 = 20 x 2 + 10
20 = 10 x 2 + 0
So, the HCF of 120 and 50 is 10.
∴ The fruitseller can make stacks of 10 mangoes and 10 apples

Fundamental Theorem of Arithmetic : 
Every composite number can be expressed ( factorised ) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.

A number s is called irrational if it cannot be written in the form p/q where p and q are integers and q ≠ 0

Rational Numbers and Their Decimal Expansions
Rational numbers have either a terminating decimal expansion or a non-terminating repeating decimal expansion.

Real number is a rational number of the form p/q where the prime factorisation of q is of the form 2n5m and n, m are some non-negative integers.

Quadratic Equation

A quadratic equation in the variable x is an equation of the form ax2 + bx + c = 0, where a, b, c are real numbers, a ≠ 0. For example x2 + 4x + 4 = 0, x2 - 9 = 0

Standard form of a quadratic equation.
ax2 + bx + c = 0, where a, b, c are real numbers, a ≠ 0

In general, a real number α is called a root of the quadratic equation ax2 + bx + c = 0, a ≠ 0 if a α2 + bα + c = 0. We also say that x = α is a solution of the quadratic equation, or that α satisfies the quadratic equation. 

The zeroes of the quadratic polynomial ax2 + bx + c and the roots of the quadratic equation ax2 + bx + c = 0 are the same.

Method of completing the square
example 2x2 – 7x + 3 = 0

2x2 – 7x = - 3
On dividing both sides of the equation by 2, we get
x2 – 7x/2 = -3/2
x2 – 2 * x * 7/4 = -3/2
On adding (7/4)2 to both sides of equation, we get
(x)2 - 2 * x * 7/4 + (7/4)2 
= (7/4)2 - 3/2
(x - 7/4)2 = 49/16 - 3/2
(x - 7/4)2 = 25/16
(x - 7/4) = ± 5/4
x = 7/4 ± 5/4
x = 7/4 + 5/4 or 
x = 7/4 - 5/4
x = 12/4 or x = 2/4
x = 3 or 1/2

Method of finding roots by quadratic formula
2x2 – 7x + 3 = 0

On comparing this equation with ax2 + bx + c = 0, we get
a = 2, b = -7 and c = 3
By using quadratic formula, we get
x = -b ± √b2 - 4ac/2a
x = (7±√49 - 24)/4
x = (7±√25)/4
x = (7±5)/4
x = 7+5/4 or 
x = 7-5/4
x = 12/4 or 2/4
∴ x = 3 or 1/2

Nature of roots
We know that the roots of the equation ax2 + bx + c = 0 are given by
quadratic formula,
x = -b ± √b2 - 4ac/2a

If b2 – 4ac > 0, we get two distinct real roots
-b + √b2 - 4ac/2a

and
-b - √b2 - 4ac/2a

If b2 – 4ac = 0, then x = -b/2a
We can say that the quadratic equation ax2 + bx + c = 0 has two equal real roots in this case

If b2 – 4ac < 0, then there is no real number whose square is b2 – 4ac.
∴ There are no real roots for the given quadratic equation in this case.

Since b2 – 4ac determines whether the quadratic equation ax2 + bx + c = 0 has real roots or not, b2 – 4ac is called the discriminant of this quadratic equation.

A quadratic equation ax2 + bx + c = 0 has
(i) two distinct real roots, if b2 – 4ac > 0,
(ii) two equal real roots, if b2 – 4ac = 0,
(iii) no real roots, if b2 – 4ac < 0.

A quadratic equation in the variable x is an equation of the form ax2 + bx + c = 0, where a, b, c are real numbers, a ≠ 0. For example x2 + 4x + 4 = 0, x2 - 9 = 0

Standard form of a quadratic equation.
ax2 + bx + c = 0, where a, b, c are real numbers, a ≠ 0

In general, a real number α is called a root of the quadratic equation ax2 + bx + c = 0, a ≠ 0 if a α2 + bα + c = 0. We also say that x = α is a solution of the quadratic equation, or that α satisfies the quadratic equation. 

The zeroes of the quadratic polynomial ax2 + bx + c and the roots of the quadratic equation ax2 + bx + c = 0 are the same.

Method of completing the square
example 2x2 – 7x + 3 = 0

2x2 – 7x = - 3
On dividing both sides of the equation by 2, we get
x2 – 7x/2 = -3/2
x2 – 2 * x * 7/4 = -3/2
On adding (7/4)2 to both sides of equation, we get
(x)2 - 2 * x * 7/4 + (7/4)2 
= (7/4)2 - 3/2
(x - 7/4)2 = 49/16 - 3/2
(x - 7/4)2 = 25/16
(x - 7/4) = ± 5/4
x = 7/4 ± 5/4
x = 7/4 + 5/4 or 
x = 7/4 - 5/4
x = 12/4 or x = 2/4
x = 3 or 1/2

Method of finding roots by quadratic formula
2x2 – 7x + 3 = 0

On comparing this equation with ax2 + bx + c = 0, we get
a = 2, b = -7 and c = 3
By using quadratic formula, we get
x = -b ± √b2 - 4ac/2a
x = (7±√49 - 24)/4
x = (7±√25)/4
x = (7±5)/4
x = 7+5/4 or 
x = 7-5/4
x = 12/4 or 2/4
∴ x = 3 or 1/2

Nature of roots
We know that the roots of the equation ax2 + bx + c = 0 are given by
quadratic formula,
x = -b ± √b2 - 4ac/2a

If b2 – 4ac > 0, we get two distinct real roots
-b + √b2 - 4ac/2a

and
-b - √b2 - 4ac/2a

If b2 – 4ac = 0, then x = -b/2a
We can say that the quadratic equation ax2 + bx + c = 0 has two equal real roots in this case

If b2 – 4ac < 0, then there is no real number whose square is b2 – 4ac.
∴ There are no real roots for the given quadratic equation in this case.

Since b2 – 4ac determines whether the quadratic equation ax2 + bx + c = 0 has real roots or not, b2 – 4ac is called the discriminant of this quadratic equation.

A quadratic equation ax2 + bx + c = 0 has
(i) two distinct real roots, if b2 – 4ac > 0,
(ii) two equal real roots, if b2 – 4ac = 0,
(iii) no real roots, if b2 – 4ac < 0.

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