MATHEMATICS
…………………………………………………………………………………………….1
MODULE 1
Contents
CLASS 1
NUMBERS
Class Assignment 1
Home Assignment 1
2
5
5
CLASS 2
RATIONAL NUMBERS
Class Assignment 2
Home Assignment 2
6
7
8
CLASS 3
REAL NUMBER
Class Assignment 3
Home Assignment 3
9
11
11
CLASS 4
ABSOLUTE VALUE OF REAL NUMBER
Class Assignment 4
Home Assignment 4
12
15
15
CLASS 5
SQUARES, SQUARE ROOTS OF NUMBERS
Class Assignment 5
Home Assignment 5
16
19
20
CLASS 6
EXPANSIONS & IDENTITIES
Class Assignment 6
Home Assignment 6
21
22
22
CLASS 7
QUADRATIC EQUATIONS
Class Assignment 7
Home Assignment 7
23
23
24
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MATHEMATICS- MODULE 1
CLASS 1
1.0.1 NUTURAL NUMBERS
The numbers 1, 2, 3, …. are called the natural numbers. This is represented using the set notation
N = {1, 2, 3 ….}.
The set of natural numbers exhibit the following properties:
(1) We can add, multiply any two natural numbers and the resulting number is also natural number.
(2) The other two operations – subtraction and division cannot be done in the set of natural numbers.
Because 2 – 3 = - 1 but – 1  N. Similarly 2  3 =
2
3
1.0.2 WHOLE NUMBERS
If 0 is added to N then we get a set {0, 1, 2, ….}. This is called a set of whole numbers.
1.0.3 INTEGERS
We get another set called set of integers (Z) by adding all the negative integers to the set of whole numbers i. e. Z
= {…-3, -2, -1, 0, 1, 2, 3 …….}
1.1 OPERATION ON INTEGERS : Fundamental Operation on Integers
Addition: Using Number Line:
(i) (+ 1) + (+ 5)
-6
-5
-4
-3
-2
-1
0
+1
+2
+3
+4
+5
+6
From the number line (+1) + (+5) = + 6.
(ii) (–2) + (–3)
-6
-5
-4
-3
-2
-1
0
+1
+2
+3
+4
+5
+6
+2
+3
+4
+5
+6
+2
+3
+4
+5
+6
From the number line (–2) + (–3) = – 5
(iii) (3) + (– 1)
-6
-5
-4
-3
-2
-1
0
+1
From the number line (3) + (–1) = 2
(iv) (– 4) + (2)
-6
-5
-4
-3
-2
-1
0
+1
From the number line (–4) + (2) = – 2
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MATHEMATICS- MODULE 1
Rules:
1) If add two positive integers, find their sum and give this sum the positive sign.
2) To add two negative integers, find their sum, and give this sum a negative sign.
3) To add integers with opposite sign, find the difference between their absolute values and give the
difference, the sign of integer having the greater absolute value.
Conclusion

The sum of two positive integers is positive.

The sum of two negative integers is negative.

The sum of a positive integer and a negative integer is positive if the greater integer (absolute
value) is positive and is negative. If the greater integer (absolute value) is negative.
Multiplication of Integers
To multiply 2 integers, first find the product of the two numbers and the sign of the product may be
assigned as follows 
if both integers are positive, the product is positive.

If both integer are negative, the product is positive, because (–) x (–) = (+).

If one integer is positive, another negative, then the product is negative.
Division of Integers:
The division of 2 integers is same as division of other numbers, but the sign may be assigned as below:

If a positive integer is divided by a negative integer or a negative integer is divided by a negative
integer in both cases, the quotient is negative.

If a negative integer is divided by a negative integer, or a positive integer is divided by a positive
integer, the quotient is positive.
1.2 SOME IMPORTANT PROPERTIES OF NUMBERS
Consider the set of integers Z = {…-3, -2, -1, 0, 1, 2, 3, …}
Here the set {…-2, 0, 2, 4, …} is called set of even numbers, the set
{…-3, -1, 1, 3, …} are called set of odd numbers.
Even numbers are multiple of 2 i.e. they are exactly divisible by 2. Where as odd numbers are not
divisible by 2.
Properties

Sum (difference) of two even numbers is an even number.

Product of two even numbers is an even number.

Sum (difference) of two odd numbers is an even number.
 Product of two odd numbers is an odd number.
You may remember these properties using following table.
Sum
Even
Odd
Even
Even
Odd
Odd
Odd
Even
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MATHEMATICS- MODULE 1
Product
Even
Odd
Even
Even
Even
Odd
Even
Odd
Some important properties of number are listed below:
1) Closure property - Sum and product of two natural numbers is a natural number.
i.e., a, b  N, a + b = c  N.
and  a, b  N, a x b = c  N.
Closure property can be checked for other sets also
 a, b  W, a + b = c  W
and
 a, b  W, a x b = c  W
and,  a, b  Z, a + b = c  Z
 a, b  Z, a – b = c  Z
 a, b  Z, a  b = c  Z
Therefore, we can say Natural numbers and whole Numbers are closed with respect to addition and
multiplication and not closed with respect to subtraction and division. Integers are closed with
respect to addition, subtraction and multiplication and closed with respect to division.
2) Commutative Property
 a, b  N, a + b = b + a
 a, b  w, a + b = b + a.
 a, b  Z, a + b = b + a
and
and
and
a  b = b  a.
a  b = b  a.
a  b = b  a.
Therefore, we can say Natural numbers, Whole numbers; Integers satisfy commutative property
under addition and multiplication.
3) Associative Property
 a, b, c  N, (a + b) + c = a + (b + c)
and
(a  b)  c = a  (b  c)
 a, b, c  W, (a + b) + c = a + (b + c)
and
(a x b) x c = a  (b  c)
 a, b, c  Z, (a + b) + c = a + (b + c)
and
(a x b) x c = a  (b  c)
Natural numbers, Whole Numbers, Integers satisfy associative property under addition and
multiplication.
Some Important Results to remember
1) The sum of two natural numbers is also a natural number
2) The product of two natural numbers is also a natural number.
3) The difference of two natural numbers, division of one natural number by other need not always be
a natural number.
4) The sum product of two whole numbers gives whole number.
5) The difference of two whole numbers, division of one whole number by the other need not always
give a whole number.
6) ‘0’ is the identity elements for addition.
7) ’1’ is the identity element for multiplication.
8) The sum product difference of two integers give an integer.
9) The division of two integers need not always give an integer
10) Also we can have two more sets, one of positive integers and another, of negative integers.
Z+ = {+ 1, + 2, + 3, …}, Z– = {– 1, – 2, – 3, …}
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MATHEMATICS- MODULE 1
CLASS ASSIGNMENT- 1
1. Find the sum of
(i) (– 3) and (10)
(ii) (– 5) and (– 2)
(i) Hint: when we are adding positive number to a negative number we have to go towards zero
same number of steps that of a positive number on number line.
(ii) Hint: do the addition like positive numbers and then place a negative sign.
2. Simplify :
(i) 6 + (– 4) + (– 2)
(ii) (– 28) + 27 + 2
(i) hint: first add both the negative numbers. Then add the resultant negative number to the given
positive number. [ Or]
start solving the problem from left side ie.
6+(-4). Then add (-2)
(ii) Hint: follow the same method given above. Ie. Add positive numbers first and then the negative
number
3. Decide which integer is larger.
(i) 10 and – 30
(ii) – 9 and – 6
4. Find the product of
(i) 5 x (– 1)
(ii) (– 28) x (– 3)
5. Simplify
(i) (– 5) x (10) x (– 4) x (– 7)
(ii) (15) x (– 3) x (2) x (– 2)
Hint: when there are odd no. of negative signs we get a negative sign in the product.
6. Divide
 25
(i)
5
(ii)
 72
3
(ii)
( 10) x (4) x ( 11)
( 8) x (5)
7. Simplify
(i)
( 8) x ( 3)
( 2)
Hint: Ignore the signs in the problem. Calculate the numerator and denominator separately.
then divide. Now if there are odd number of negative signs in the problem put negative sign,
otherwise positive..
8. If a  N then a2 + a  N. (T/F)
Hint: Take any natural number check the rule. Ie. 22 +3=7 this is also a natural number.so the rule.
9) On a road, the houses on either side are directly opposite to one another. Starting from one side of
one end of the road, houses are numbered 1, 2, 3…… On reaching the other end of the road
numbering is continued on the other side of the same end. If the houses numbered 25 and 40 are
opposite to each other, the number of houses on the road is: [NTSE NOVEMBER 2006, KAR]
a) 60
b) 62
c) 64
d) 68
The figure shows how numbering was done
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MATHEMATICS- MODULE 1
1
2
3
-
-
-
10
9
8
7
Here the numbers 3 and 10 are opposite to each other. Total numbers are 12. Now imagine the houses
are numbered in the same manner and calculate.
HOME ASSIGNMENT – 1
(Home assignments should be submitted in a separate sheet before the next class)
1. Find the sum of
(i) (9) + (– 6)
(ii) (– 2) + (– 17)
(iii) (– 5) + (10)
2. Find the sum of 3 + (– 5) + 4 using the number line.
3. Decide which integer is larger.
(i) (– 5) and 0
(ii) (– 12) and (– 13)
(iii) (100) and (– 500)
Hint: 3 is smaller than 4 but -3 is bigger than -4.
4. Find the product of
(i) (– 100) x (– 41)
(ii) (8) x (36)
(iii) (14) x (– 5)
5. Simplify
(i) (6) x (– 6) x 9
(ii) (– 4) x (– 3) x (– 2) x (– 9)
(iii) (8) x (– 5) x (– 3)
6. Divide
(i)
 105
3
(ii)
1441
 11
(iii)
 9999
9
7. Simplify
(i)
( 21) x ( 9)
(3)
(ii)
( 75) x (5) x (6)
(5) x ( 5)
(iii)
(76) x (8) x (2)
(4) x (4)
8. Show that for every natural number n, the number n2 + n + 1 is an odd number. (Hint : n2 + n + 1 =
n(n + 1) + 1)
Hint: the product of any two consecutive numbers is always even.
9. Let N be the set of natural numbers and let M = {x where x  N and 10 < x2 < 50}. Then exhibit M in
the tabular form.
Hint: take the approximate square root of 10 and 50. The numbers in the set are between these two
numbers.
10. If the sum of 2 natural numbers is not divisible by 5, then at least one of the numbers is not divisible
by 5. (T/F)
11) The numbers 1 to 100 are written, from left to right in a raw as a single number. The 75 th digit
starting from the left in this new number is
a) 1
b) 2
c) 3
d) 4
Hint:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 . . . . . .
Single digit
Double digits
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MATHEMATICS- MODULE 1
75 – 9 = 66
there are 66 digits
ie, 33 numbers. Count 33 numbers from 10.
12) Rama is 8 cm taller than Krishna. Hari is 12 cm shorter than Rama. If Krishna is 125 cm in
height. How tall is Hari?
a) 129
b) 121
c) 105
d) 113
Hint: krishna’s height is given. Calculate rama’s height first.
CLASS - 2
1.4 RATIONAL NUMBERS
p
, where
q
p, q  Z, q  0, are called Rational numbers. The rational number set is represented using the letter Q.
p
i.e., Q = {x | x = , p, q  Z, q = 0}.
q
2
3
Example: ,  etc.
3
5
 Set of rational numbers is denoted by Q
Rational Number: The number which can be written as a fraction or the number of the form
 Every natural number is a rational number but a rational number need not be a natural number.
 Zero is a rational number.
5
where p = 5 and q = 1
1
 Every integer is a rational number but a rational number need not be an integer.
 All fractions are rational numbers.
 Every integer is also a rational. For example 5 =
 Every fraction is a rational number but a rational number need not be a fraction.
 The addition, subtraction, product, division (except by 0) of two rational numbers results in a rational
number.
 Any rational number can be expressed in decimal form.
For example
1
1
1
= 0.5,
= 0.333,
= 0.125
2
3
8
1.4.1 OPERATIONS ON RATIONAL NUMBERS
Example:
Add
5
1
and
2
3
 Since the denominators are different. We have to make a common denominator (by taking LCM = 6)
So
5 15
=
2
6
So
5 1 15 2 17
 
 
2 3 6 6 6
and
1 2
=
3 6
The same may, subtraction can be carried out.
6
1
and
5
5
6 1
6 1
6
 =
 =
5 5
5 5
25
1
2
Example:
Divide  7 by
3
5
Example:
Multiply
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MATHEMATICS- MODULE 1
1  36 

 7 

5
5 

 36 2
 36 3
 18  3  54

 =
 =

5
3
5
2
5
5
EQUALITY OF RATIONAL NUMBERS
Example: Are the rational numbers
Method 1
8  2 2 2  2


12
3 2  2
3
8
 50
and
equal?
12
75
 50  5  2  5  2


75
5 3 5
3
Method II
 8  50
=
12
75
Cross multiply
75  (-8)
,
-600
,
So, they are equal.
So, they are equal.
8
12
 50
75
-50 x 12
-600
CLASS ASSIGNMENT - 2
1. Separate positive and negative rational numbers from the following :
 3 1
5 10
 11 7
,
, 0, ,
, 104,
,
5
8
1 3
10  2
2. Which of the following rational numbers are positive?
2
6
17
a)
b)
c)
3
5
4
3. Express
d) – 101
5
as a rational number with denominator 168.
 21
Hint: divide 168 with -21 and multiply the fraction with that quotient.
4. Write five rational numbers between
1
2
and
10
10
Hint: multiply 1/10 and 2/10 with 10/10. write the Rational numbers between the two numbers you
got.
5.
1 4 1 3 
     is equal to [NTSE NOVEMBER 2010, KAR]
15  15 3 45 
a)
45
4
b)
1
8
c)
8
3
d)
4
45
Hint: make denominators same(45) for all the numbers and calculate.
1 1 1
  =
 2 4 3
6. 6. 
a) 5
24
[NTSE NOVEMBER 2006, KAR]
b) 10
12
c)  1
12
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d) 5
12
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CFAL
MATHEMATICS- MODULE 1
Hint: take LCM and calculate.
HOME ASSIGNMENT - 2 .
(Home assignments should be submitted in a separate sheet before the next class)
1. Write down the rational number whose numerator is (– 3) x 4, and whose denominator is (34 – 23) x
(7 – 4).
2. Which of the following rational numbers are positive?
 15
2
b)
c) – 12256
7
9
1
3. If
is an integer. What are the possible values of x?
x
Hint: ½ is not an integer. Therefore x should be less than 2.
a)
d)
 11
4
4. Every rational number must be a whole number. (T/F)
5. The number 0 is both positive and negative. (T/F)
6. Simplify
.
1
2
2
4
2
5
2
2
6
7
Hint: start the calculation from the bottom most denominator and go up step by step.
6
First calculate 2 + ---7

3
7. Express
as a rational number with denominator 125.
5
Hint: 125/5=25 multiply both numerator and denominator with 25
 1  1
4   2 
8. Simplify  7   4  
 1  1
 3   1 
 2  7
1
.
1
2
1
2
5
1
5
9. The rational number between 1 and 1 is
2
a)
2
5
b) 1
5
3
c) 3
5
d)
4
5
Hint: convert the given fractions into decimal numbers. Write the fraction between these decimal
numbers.
2
3
a) Lies to the left side of 0 on the number line
b) Lies on the right side of 0 on the number line
c) It is not possible to be represented on the number line
d) Cannot be determined on which side the number lies on the number line
10. A rational number
Hint: convert the fraction into decimal
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MATHEMATICS- MODULE 1
11. The difference between the greatest and least numbers of
a) 2
b) 4
c) 10
9
9
9
5
9
,
1 , 11
9 9
is
d)
2
3
Hint: higher the numerator greater the value.
11 5
= ________

39 26
a) 149
b) 1 + 71
39
78
12. 2 
c) 149
d) 149
76
98
hint: take LCM and calculate.
13. What number should be added to  5 so as to get 3 ?
6
a)
7
3
b)
1
2
3
c)
2
8
3
d)  8
3
Hint: let x/1 be the fraction to be added to -5/6.
CLASS - 3
1.5 REAL NUMBERS
Real Numbers include all the numbers. Whole numbers, natural numbers, integers, fractions, rational
numbers and also IRRATIONAL NUMBERS. Any numbers that can be plotted on the number line are
called real numbers.
1.5.1 IRRATIONAL NUMBERS:
The numbers which are not rational are called irrational numbers. i.e. the numbers which cannot be
p
written in the form . Where p and q are integers)
q
2,
Ex:
7,
 etc.
Note:
 Both Rational and Irrational Numbers can be represented in decimal form also.
 Decimals can be
6
1
= 0.5,
= 1.2) / Non – Terminating
2
5
4
3
0.1111…,
= 0.428571…,
= 0.121212…)
33
7
 Terminating (For Example,
 Recurring (For Example, 0.313131…, 0.55555) / Non-Recurring
11.1872931275…)
(For Example,
1
=
3
(For Example,
 Every rational number can have
 Terminating decimal expression
 Recurring decimal expression
 Every irrational number can have
 Non-terminating decimal expression
 Non-recurring decimal expression
Where as irrational numbers have non-terminating and non-recurring decimal expression. This is the
difference between rational and irrational number.
Example 1: Classify the following into rational and irrational decimal expressions:
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MATHEMATICS- MODULE 1
i) 0.1111…,
ii) 0.237698782…,
iii) 0.15625,
iv) 1331.133113311331…, 7158634281…, v) 715.86
vi)
22
7
vii) 3.1415…..
 Rational Numbers :


0.11111…,
1331.13311331
22
7



0.15625
715.86
 Irrational Numbers:


1.5.2
0.23769872…
3.1415 …..
REPRESENTATION OF NUMBERS ON THE NUMBER LINE
1) Draw the number line and plot only the integers
-4
-3
-2
-1
0
1
2
3
4
5
2) Between 1 and 2, there are infinitely many points so the point right in midway between 1 and 2 is the
rational number
1
2
-1
1
0
½
Irrational number: consider the number line given below:
P
√2
1
O
1
A
B
Let OA = 1 unit
At A draw PA perpendicular to the line.
Let AP = 1 unit
Then, OP =
12  12  2
With OP as radius and O as centre, if we draw an arc cutting the number line at B then, OB =
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2 units
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MATHEMATICS- MODULE 1
Now,
2 is not in the form
p
. This type of numbers is called irrational numbers.
q
Thus, a number, which cannot be represented in the form
p
is called irrational number, and is denoted
q
by Q'
Thus, the number line will be complete if we take,
1. Integers (both positive and negative)
2. Rationals ( in the form p/q)
3. Irrationals (not in the form p/q)
Note : We have thus a complete system as shown below:
NWZQ
And
Q  Q' = R
CLASS ASSIGNMENT - 3

1)

4  9 is
A) an irrational number
B) A fractional number
C) an integer
D)
Hint:
4 and
9 cannot be multiplied
4  9  36
2) Addition of two irrational number gives an irrational number only. State True / False
Hint; let us
a  b and
take two irrational numbers
b add. you get ‘a’ which is
rational.number.
3  2 2 is
3)
A) an irrational number
B) A fractional number
5 2
C)
D) an integer
Hint; can the given number is represented in p/q form?
4) 0.333333333 … = p
q
,
p
q
is [NTSE NOVEMBER 2008, KAR]
a) 1
b) 2
c) 3
d) 1
3
6
9
33
Hint: choose the fraction that gives the given decimal number and is in its simplest form from the
answers
5) How many numbers from 1 to 100 have digit 5 (at least one) in them?
A) 10
B) 15
C) 11
D) 20
Hint: write the multiples of 5 that comes from 1 to 100
6) If 10 = 3.3333333 …, then what is 10
3
4

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MATHEMATICS- MODULE 1
A) 4.4444…
B) 0.44444…
C) 2.5
D) 10 is irrational
4
Hint: divide and write
7) The number
a)
1 lies between [NTSE MAY 2009, NAT]
1 5
1
1
and
3
2
b) 1 and
2
Hint: the value of
1
c)
2
1
1
and
4
3
d)
1
1
and
5
4
5 is slightly more than 2. So 1+ 5 is again slightly more than 3.
1
1
1
1




2
2
2
2
8)
b) 2
2
a) 2 2
c)
2
d)
1
4 2
2,
Hint:1: take LCM
Hint 2: 2= 2  2
1
1
1
a 
a = (given that a 1) [NTSE MAY 20011, NAT]
1 a
1 a
a
9)
a)
1

a
1
a
b) 0
c)
2
a
d)
1
2 a
1
1
1
a 
a
1 a
1 a
1
1
a
Hint 1:
Hint 2: first calculate numerators taking LCM . then reciprocate the denominator.
10) Which of the statements is false?
A) 2 + 3 >
C)
Hint:
2 3
1
1

23
23
23
B) 2 + 3 >
D)
3
9

1
3
a  b  a  bor a  b and 1/a<a
HOME ASSIGNMENT - 3
(Home assignments should be submitted in a separate sheet before the next class)
1. Can a fraction
x
be rational if x and y are irrational?
y
2. Can the sum of x + y be rational if x is rational and y is irrational?
3. Can a product x and y be rational if one of the factor is rational and the other is irrational?
4. Can the difference x – y be rational if x is rational and y is irrational?
5. Can the fraction
x
be rational if x is rational and y is irrational?
y
6. Classify the following into rational and irrational decimal expressions :
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CFAL
MATHEMATICS- MODULE 1
12.121212…, –156.7649, 0.56725672…, 36.36, 712.547984681…, – 98.764379…
7. Check whether the given numbers are co-primes: .
a) 56 and 1331
b) 80 and 900
c) 71 and 85
Hint: find those numbers whose HCF is 1.
8. Which fraction is the smallest?
41 32 54
,
,
52 39 65
Hint: cross multiply first and second fractions, and first and third fractions. then compare
9. The decimal equivalent to 7 is _______.
12
Hint; divide and write
10. The decimal equivalent to 49 is _______.
396
3
x
then x is

x
 27
a) A rational number
b) Not a rational number
c) An integer
d) A natural number
Hint: cross multiply and solve for x.
11. If
12. Set of Natural numbers is a subset of
a) Set of even numbers
b) Set of odd numbers
c) Set of composite numbers
d) Set of real numbers
Hint: set of real numbers includes positive, negative, rational and irrational numbers along with
zero.
13. Represent
3 on the number line.
CLASS 4
2.0 ABSOLUTE VALUE OF REAL NUMBER
If a number is a real number then, absolute value of ‘a’ is written as |a| and is always non-negative.
a if a  0

i.e., | a | 0 if a  0

 a if a  0
Example: |5| = 5 because, 5 > 0
|–5| = –(– 5) = 5 because – 5 < 0
Example 1: Find the solution of the equation |x – 5| = 3
Solution:
When, x – 5 > 0, |x – 5| = x – 5 = 3  x = 8
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MATHEMATICS- MODULE 1
CFAL
When, x – 5 < 0, |x – 5|= – (x – 5) = 3  – x + 5 = 3  – x = – 2  x = 2
The solution of the equation is 2 and 8.
Consider number 12, when we divide 12 by 2, we get remainder 0,
We can write:
12 = 2  6 + 0
=26
Hence we call 2, a factor of 12. Thus a number ‘a’ is a factor of another number ‘b’ if a divides b
completely (i.e. without leaving reminder).
And we can write b = k.a when kN
Common factor: A factor which divides two or more numbers is called a common factor.
For example, 2, 4, 8 are the common factors of 16 and 24 ; x is a common factor of 2x, 3x and 4x.
Note:
1. For any number, at least two factors are there i.e. the number itself and 1.
2. If a number has only two factors then it is called prime number e.g. 5 = 51, and hence 5 is a prime
number.
3. If a number ‘a’ has a factor other than ‘a’ then it is called composite number. Example: 6 = 2  3 and
2, 3 are factors of 6, hence 6 is a composite number.
4. If two numbers have a common factor, it will divide the sum and difference of the two numbers.
Example 2: Find all the factors of 18.
Solution: Consider 18 = 1  2  32
Factors of 18 are 1, 2, 3, 6, 9, 18, i.e. there are 6 factors.
2.1 PRIME AND COMPOSITE NUMBERS
2.1.1 Prime number
A number, which cannot be divisible by any other number except 1 and itself, is called prime number.
Example: 2, 3, 5, 7, … are prime number.
2.1.2 Composite number
A number, which is not prime, is called composite number.
Example: 4, 6, 8, 10 etc are composite numbers.
Co-prime Numbers: Two numbers are co-prime to one another if they have no factors common
except unity. For Example,15 and 28.
Note:
1. By convention 1 is not a prime number.
2. 2 is the only even prime number.
3. All composite numbers can be written as product of prime factors.
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2
Example 2: 4 = 2 where 2 is prime
12 = 4 x 3 = 22 x 3 where 2 and 3 are prime etc.
The natural number can be divided into two sets as follows:
Prime
2, 3, 5…
NATURAL NUMBER
Composite (non – primes)
1, 4, 6….
Example 3: Express 200 as a product of prime numbers.
Solution: 200 = 2 x 100 = 2 x 10 x 10 = 2 x 2 x 5 x 2 x 5 = 23 x 52
Example 4: Show that the product of two consecutive integers is always an even number.
Solution: consider P = n(n + 1)
Case 1: If n is even, n + 1 is even. Then P = n(n + 1) = even x odd = even.
Case 2: If n is not even (n is odd), then n + 1 is even.
Then P = n(n + 1) = odd x even = even.
 P = n (n + 1) is always even.
2.2
TEST OF DIVISIBILITY
Divisibility by 2: all even numbers are divisible by 2.
Example: 24362 is divisible by 2 as it is an even number.
Divisibility by 3: if the sum of all digits of a number is divisible by 3 then the number is divisible by 3.
Example: 367 is not divisible by 3. Because 3 + 6 + 7 = 16 which is not divisible by 3.
Divisibility by 4: a number is divisible by 4 if the last two digits are divisible by 4.
Example: 476 is divisible by 4. Because 76 is divisible by 4.
The number 466 is not divisible by 4 because 66 is not divisible by 4.
Divisibility by 5: a number is divisible by 5 if the last digit is 5 or 0.,
Example: 125, 360, etc are divisible by 5.
Divisibility by 6: An even number whose sum of digits is divisible by 3 is divisible by 6.
Example: 540 is an even and is divisible by 3 hence will be divisible by 6.
Divisibility by 8: A number whose last three digits are divisible by 8 is divisible by 8.
Example: 864, 17888 etc.
Divisibility by 9: If the sum of all digits of a number is divisible by 9, then the number is divisible by 9.
Example: 325647 is divisible by 9.
Because 3 + 2 + 5 + 6 + 4 + 7 + = 27 which is divisible by 9.
Divisibility by 11: A number is divisible by 11, if the difference between the sum of the digits in the
even places and sum of the digits in the old places is either 0 or a number divisible by 11 (or a multiple
of 11).
Example: 45265 is divisible by 11, because 4 + 5 + 2 = 11 and 5 + 6 = 11 and 11 – 11 = 0.
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MATHEMATICS- MODULE 1
CLASS ASSIGNMENT - 4
1. Find the factors of 12.
2. Separate the prime and composite numbers :
1001, 1759, 1134, 5281, 3,
125
.
5
3. Find the factors of the following :
a) 91
b) 1331
c) 60
4. Which of the following is not possible? Why?
|x|2 = x2
|x2| = – x2
State which of the following are true and which are false. Give reasons
5. Every odd perfect square can be written in the term 8k+1 where k is some integer.
Hint; some odd perfect squares: 32 =9, 52 =25,……..when divided by 8 gives 1 as remainder.
6. If a < b, then always |a| < |b|.
Hint: for example take a=-1,b=1 or any other numbers and apply the rule.
7. The number that is neither prime nor composite is
a) 0
b) 1
c) 2
8. Addition of a prime number with an odd number always gives
a) Odd number
b) Even number
c) Prime number
d) 3
d) None of these
Hint; randomly select some prime numbers and check the rule.
9. How many prime numbers smaller than 100 leave the remainder 3 when divided by 4?
[NTSE NOVEMBER 2007, KAR]
a) 11
b) 12
c) 13
d) 14
Hint: when a prime number is divided by 4 it gives a remainder 3. So
Prime number = 4k+3.
10. What is the value of - -  -  a    ? (Where x represents absolute value of x)
a) a
b) 0
c) –a
d) None of these
Hint: the absolute value of a negative numbers is always positive.
HOME ASSIGNMENT – 4
(Home assignments should be submitted in a separate sheet before the next class)
1. Express 169 as a product of prime numbers.
Hint: a number only two factors ie; 1 and the number itself is called prime number.
2. It is given that n, n + 2 and n + 4 are primes. Find n.
3. Separate the prime and composite numbers :
25
111
, 1005, 113, 1,
5
3
4. Find x if
a) |x + 2| = 7
234,
b) |2x – 1| = 13
Hint: x+2=7 or x+2=-7 solve for x.
5.
9
3 4
=
 
 17 7 5
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MATHEMATICS- MODULE 1
a)
117
595
b)
 117
595
c)
595
117
d) None of these
Hint:1) first calculate the absolute value by taking LCM.
(2)  -  is positive
6. State True or False
i) If one of the numbers 2n – 1 and 2n + 1, where n  2 is prime then the other is a composite
number.
ii) If two numbers have a common factor, then it will divide their sum and difference exactly
Hint: Let the numbers be c and d
c = ak
(where k is the common factor)
d = bk
Now we have to check whether the common Factor divides the sum and the difference.
7. The number of digits in the number N = 212  58 is –
a) 9
b) 10
c) 11
d) 12
Hint write the given number in powers of 10 ie; 10( )x24
8. The three digit number 2a3 is added to the number 326 to give 5b9. If 5b9 is divisible by 9, a + b =
a) 2
b) 4
c) 6
d) 8
Hint: 1) apply the divisibility rule for 9 to the number 5b9
2) 5+b+9=18 solve for b.
9. A book has pages numbered from 1 to 192 (total of 96 sheets). Some 25 sheets are pulled out of it
at random. Then, the sum of these 50 numbers (on the 25 sheets) cannot be:
a) 1001
b) 1567
c) 2008
d) 3003
Hint: sum of the numbers in 25 sheets is a odd number (oddxodd)
sum of the numbers in 96 sheets is a even number (oddxeven)
difference in sum is even number – odd number=odd number.
10. This year, my age is a multiple of 7. Next year, it will be a multiple of 5. I am above 20 years but
less than 80. What is my age?
a) 35
b) 49
c) 56
d) 70
Hint: from the list of multiples of 7 greater than 20, choose the one which when 1 is added becomes
a multiple of 5.
11. The sum of two consecutive odd numbers is 2004, then the smaller of the two numbers could be,
a) 2001
b) 1001
c) 1003
d) 3
Hint: n+(n+2)=2004 solve for n
12. A six digit number of the form abcabc is always divisible by
a) 2
b) 11
c) 3
d) 6
Hint: (a+c+b) - (b+a+c)=0
13. 135 is a natural number whose digit product is (1  3  5) 15. How many such 3 digit numbers have
digit product equal to 12?
a) 4
b) 9
c) 3
d) 6
Hint: prime factorize 12. And form the numbers with those factors
14. A book with 480 pages is numbered in the usual way. How many digits will the numbering need?
a) 1332
b) 1500
c) 4800
d) 1234
Hint: there are 9 one digit numbers, 90 two digit numbers and 381 three digit numbers are needed.
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15. November 8, 1988 is a peculiar day. If we write it as 8/11/88, we have 8  11 = 88. How many
such days are there in 1972? [NTSE NOVEMBER 2006- KAR]
Hint : write the factors for 72. And the numbers of dates possible.
16. A number when divided by 195 leaves a remainder 47. If the same number is divided by 15, then
remainder will be [NTSE MAY 2008- NAT]
a) 1
b) 2
c) 3
d) 4
Hint: Let N be the number.
According to the given information:
N 195k  47

15
15
17. If x < -2, then 1 -  1 + x   equals - [NTSE MAY 2008- NAT]
a) 2 + x
b) x
c) -x
d) -2 –x
Hint: take x= -3 and calculate.
18. If 6xy5 is divisible by 55 then (y-x) is equal to [NTSE MAY 2011- NAT]
a) -2
b) -1
c) 1
d) 2
Hint: 6xy5 is also divisible by 11. Apply the divisibility rule.
CLASS - 5
2.3 SQUARES, SQUARE ROOTS OF NUMBERS
Squares: A number multiplied by itself gives the square of that number.
We know that, 2 x 2 = 4 i.e., 22 = 4. Therefore, we call 4 as square of 2.
Similarly, 9 is square of 3, 64 is square of 8 and so on.
Also, (– 2) x (– 2) = 4 i.e., (– 2)2 = 4.
Then, 4 is the square of either 2 or – 2 or 4 is the square of + 2 or – 2.
Square Roots: We have, 4 is a square of 2 or we can also say, + 2 is square root of 4. Also, 4 is
square root of – 2. Therefore, – 2 is square root of 4.
Hence,
4   2 , where
is the square root sign.
We say that 4 is the square of  2 or  2 is the square root of 4.
Similarly,  3 is square root of 9,  25 is square root of 625.
Note: Square root of 0 is 0 i.e.,
0 0
Square root of 1 is  1 i.e., 1  1
The positive square root of a number is known as the Principal square root of the number.
For example, + 2 is the principal square root of 4 where as – 2 is not principal square root of 4.
Similarly, + 3 is the principal square root of 9, + 10 is the principal square root of 100.
Methods to find the square root of a number
The following are some of the methods to find out the square root of a given number:

Factor Method

Division Method

Using Other Known Square Numbers
2.2.1 FACTOR METHOD
Example 1: Find the square root of the number 27225 by factor method.
Solution:
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MATHEMATICS- MODULE 1
Hence, 27225 = 5 x 5 x 3 x 3 x 11 x 11 = 52 x 32 x112
Therefore,
27225 = 5 x 3 x 11 = 165
Example 2: Examine whether 3528 is a perfect square.
Solution: To find whether the given number is a perfect square, first we shall find the square root of the
number.
3528 = 2 x 2 x 2 x 3 x 3 x 7 x 7
3528 =23 x 32 x 72
Here, since the power of 2 is odd, 3528 is not a perfect square. i.e., we get
3528  2 42
Example 3: Find all the perfect squares between 1000 and 2000, which are divisible by 13.
Solution: To have a number which is a perfect square and also divisible by 13, it must be a multiple of
132.
The only square number 32 multiplied by 132 gives a number between 1000 and 2000.
132 x 32 = 1521.
Therefore, the required square number is 1521. (1521 = 132 x 32 = (13x 3)2 = 392)
2.2.2 DIVISION METHOD
Example 1: Find the square root of 47961.
Solution: we shall take pairs of digits starting from the right extreme.
Hence,
47961  219
Square root of the numbers having decimal part
Let us consider the following examples,
(0.1)2 = 0.01
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MATHEMATICS- MODULE 1
2
(0.14) = 0.0196
(2.5)2 = 6.25
As shown in the above examples, the square of any decimal must be an even number of decimal digits.
These decimal digits are obtained by doubling the number of decimal digits in the original decimal
number.
In finding the square root of a decimal number we take pairs of digits from the extreme right, as seen in
following examples.
Example 2: Find the square root of 1218.7081
Solution: Since we have 4 decimal places in the given square number, its square root must contain
only 2 decimal places.
Now, let us pair the digits in the given number from right and find the square root.
Hence,
2.4
1218 .7081   34.91
PROPERTIES OF SQUARES
1. The square of the first 10 natural numbers are given in the table. Notice that none of them end in 2,
3, 7 or 8. Therefore, we can say that a number ending in 2, 3, 7 or 8 is never a perfect square.
Thus 32532, 4353, 5007 or 384898 are not perfect squares.
number
0
1
2
3
4
square
0
1
4
9
16
number
5
6
7
8
9
square
25
36
49
64
81
2. A number ending in an odd number of zeros is never a perfect square. Thus 3530, 55849000, or
38485000 are not perfect squares. Note, however, that this does not mean that all numbers ending
in an even number of zeros are always perfect squares. They may or may not be perfect squares.
Thus 500 is not a perfect square, whereas 400 is a perfect square.
3. Look at the squares of the first ten natural numbers in the table above. You will notice that
i) Squares of even numbers are always even.
ii) Squares of odd numbers are always odd.
This is true for all natural numbers.
4. The square of a natural number n is equal to the sum of the first n odd numbers. Thus 12 = 1
22 = 1 + 3 : sum of first 2 odd numbers
32 = 1 + 3 + 5 : sum of first 3 odd numbers
82 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 : sum of first 8 odd numbers.
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MATHEMATICS- MODULE 1
5. The difference of the squares of two consecutive numbers is equal to the sum of the numbers.
(n + 1)2 – n2 = n2 + 2n + 1 – n2 =(n + 1) + n
Thus 112 – 102 = 121 – 100 = 21 = 11 + 10
162 – 152 = 256 – 225 = 31 = 16 + 15
6. Three natural numbers a, b and c are called a Pythagorean triplet if a2 + b2 = c2
For any number n > 1, (2n, n2 – 1, n2 + 1) is a Pythagorean triplet. Thus, for n = 3, (6, 8, 10) is a
Pythagorean triplet since 102 = 100 = 62 + 82. Similarly for n = 4, (8, 15, 17) is a Pythagorean triplet
.
The square root of any proposed expression is that quantity whose square, or second power, is equal
to the given expression. Thus the square root of 81 is 9, because 92 = 81.
The square root of ‘a’ is denoted by
2
a or more simply a .
CLASS ASSIGNMENT - 5
1. Find the square root of 625.
2. Find the least factor by which 19845 has to be multiplied so that it may become a perfect square.
Hint: prime factorize the number and group.
3. Find the square roots of 3.6481 by division taking pair of digits.
4.
1764 is
a) 32
b) 42
c) 48
5) Square root of an odd number
a) need not be odd
c) even
b) odd
d) cannot be determined
6) The digit in the units place of 691874472 is
a) 7
b) 1
2
d) 38
c) 9
d) 6
2
Hint : 12 =144 digit in units place is 4. 15 =225 the digit in the unit place is 5. So the given no.
7) If 3  230   = 492a04 then a + x =
a) 4
b) 8
2
c) 12
d) 16
Hint : 32  (230 + x)2 is divisible by 9. Apply divisibility rule to 492a04 and find a, then x.
8)
How many square integers are there between 1 and 1000 that are divisible by 3? [NTSE
NOVEMBER 2006 - KAR]
a) 333
b) 31
c) 30
d) 10
Hint: square integers should be multiples of 3.
9) The smallest number by which 32 has to be multiplied to get a perfect square? [NTSE NOVEMBER
2006 - KAR]
a) 2
b) 4
c) 16
d) 6
Hint: write the smallest possible prime factors for 32
10) The number of square numbers between 1000 and 2000 is [NTSE NOVEMBER 2008 - KAR]
a) 44
b) 25
c) 13
d) 10
Hint: 302=900 and 502=2500 the numbers should be between these two.
11) Which digit is not used in first ten square numbers? [NTSE NOVEMBER 2008 - KAR]
a) 2
b) 3
c) 7
d) 8
12) How many pairs of natural numbers are there so that the difference between their squares is 60
[NTSE MAY 2011- NAT]
a) 4
b) 3
c) 2
d) 1
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MATHEMATICS- MODULE 1
check the problem once.
HOME ASSIGNMENT - 5
1. Find the square root of
(i) 16
(ii) 53361
(iii) 390625
2. Find the square root of the following numbers using long division
(i) 2
(ii) 3
(iii) 5
(iv) 7
3. The least number to be multiplied to 4563 to make it a perfect square is;
a) 27
b) 3
c) 13
d) 20
Hint: prime factorize the given number.
4. The least factor by which the number 46800 must be divided so that the quotient is a perfect
square.
a) 2
b) 13
c) 9
d) 5
Hint: prime factorize and choose a number wich does not have a pair.
5. The square root of
a) 0.45
6. If
0.081 0.484 2.5
is


0.0064 6.25 12.1
b) 0.75
x
16
=
then x =
49
49
a) 4
hint:
d) 0.99
c) 16
d) 28
c) 254
d) none of these
16
49 =4/7 do the cross multiplication.
7. the value of
a) 208
Hint:
b) 7
c) .95
99  396 is
b) 198
99  396  99  396
The smallest natural number which when added to the difference of squares of 17 and 13 gives a
perfect square is
a) 1
b) 5
c) 11
d) 24
Hint: (17+13)(17-13)=120. (2) 121 is a perfect square
9. If aabb is a four digit number and also a perfect square, then a + b = [NTSE MAY 2011 NAT]
a) 12
b) 11
c) 10
d) 9
hint: the number should be a multiple of 11 as well as multiple of 121. Like 2299 etc., (2) the
number should not end with 2,3,7,8.
11. Sum of three consecutive odd numbers is a perfect square between 200 and 400, then, the root
of this sum is
Hint: let the odd numbes be n-2,n and n+2 .solve for n.
12. n4 = 625 then n = ______
Hint: (n2)2 = 625
6561
12.
a) 9
=
b) 81
c) 3
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CFAL
MATHEMATICS- MODULE 1
13. The square root of the only square number between 1900 and 2000 is
[NTSE NOVEMBER 2009 - KAR]
a) 43
b) 44
c) 42
d) 41
14. How many square integers between 1 and 100 are divisible by 5?
[NTSE NOVEMBER 2009 - KAR]
a) 5
b) 6
c) 7
check the problem. The problem shold be 1 and 1000
L
then ‘L’ equals to [NTSE NOVEMBER 2010 - KAR]
g
15. T = 2
a)
d) 8
T2
2g
b)
T 2g
2
Hnt: square both the sides and solve for L
16. How many mm2 are there in 20 cm2 ?
a) 20
b) 200
Hint: 20cm2=10cmx2cm; 1cm=10mm
c)
T 2g
4 2
c) 2000
d)
T 2g
4
d) 20000
Class 6
EXPANSIONS & IDENTITIES
 (a + b)2 = a2 + 2ab + b2
 (a – b)2 = a2 – 2ab + b2
 (a + b)2 – (a – b)2 = 4ab,
 (a + b)2 + (a – b)2 = 2(a2 + b2)
 (a + b)3 = a3 + 3a2b + 3ab2 + b3 = a3 + b3 + 3ab(a + b)
 (a – b)3 = a3 – 3a2b + 3ab2 – b3 = a3 – b3 – 3ab(a – b)
 (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
 a 
2
2
1  a  1   2

a 
a2 
2
2
1
1
  a     a    4
a 
a

3
1   a  1   3 a  1 
 a 

a3 
a 
a 

3
 a 
3
3
1   a  1   3 a  1 


a 
a 

a3 
 (x + a)(x + b) = x2 + (a + b)x + ab
 (x – a)(x – b) = x2 – (a + b)x + ab
Example 1: Evaluate (13x – 6y)3
Solution:
(13x – 6y)3 = (13x)3 – (6y)3 – 3(13x)(6y)(13x – 6y)
= 2197x3 – 216y3 – 234xy(13x – 6y)
= 2197x3 – 216y3 – 3042x2y + 1404xy2
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CFAL
MATHEMATICS- MODULE 1
2
2
a 2
a 2
Example 2: Show that         4
2
a
2
a
Solution: Consider L.H.S.
2
2
2
2
2
2
 a  2    a  2    a    2   2 a  2    a    2   2 a  2 


 2 a   2 a   2 a 
 2  a   2   a 
 2  a 
=
a2  4  2  a2  4  2  2  2  4
4 a2
4 a2
2
= R.H.S.
2
a 2
a 2
Therefore,         4
2
a
2
a
Example 3: If a2 + b2 + c2 = 125 and ab + bc + ca = 50, find a + b + c.
Solution: we have (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
i.e., (a + b + c)2 = 125 + 2  50 = 225  a + b + c = 225 = 15.
CLASS ASSIGNMENT – 6
1. If x + y = 10 and xy = 16, find the value of x – y.
Hint: (x-y)2 = (x+y)2 – 4xy
2. If 3x + 4y = 16 and xy = 4, find the value of 9x2 + 16y2
Hint: square both the sides and solve for 9x2 + 16y2
3. Show that (3p – 5q)2 – (3p + 5q)2 + 60pq = 0
Hint: put 3p in place of ‘a’ and 5q in place of ‘b’ in the expantion of (a-b)2 and (a+b)2
4. If a2 + b2 = 13
x3 + y3 = 16
Find the value of
((ax + by ) + (ay + bx))
Check the problem
5. If x = 2p – 3q, y = p + 2q and z = q – 3p then, x2 + y2 + z2 + 2yz + 2zx + 2xy is
a) 0
b) 1
c) xyz
d) pqr
hint: taking x,y,z values calculate x+y+z. x2 + y2 + z2 + 2yz + 2zx + 2xy = (x+y+z)2 now substitute x+y+z
and calculate.
6. If x 
1  3, then, x3  1 is
x
x3
a) 9


Hint:  x 
b) 18
c) 27
d) 6
3
1
1
1

3
  x  3  3 x  
x
x
x

HOME ASSIGNMENT - 6
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CFAL
MATHEMATICS- MODULE 1
1. If x 
1  3, then x 2  1 is equal to
x
x2
a) 9
b) 7
c) 6
d) 3
c) 5
d) 30
2


Hint :  x 
1
1
2
 2 x  2
x
x
2. If x+ y = 5, x3 + y3 = 35, then x – y is equal to
a) 0
b) 1
Hint: (1) ( x – y)2 = (x + y)2 – 4xy
(2) (x + y)3 = x3 + y3 + 3xy (x + y) substitute given values and calculate xy.
3. If a + b + c = 14 and a2 + b2 + c2 = 74, then the value of ab + bc + ca is
a) 60
b) 88
c) 61
d) 0
Hint: (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) substitute the given values and calculate.
a  b2  b  c2  c  a2
4. If a + b + c = 0, then the value of
a) 0
b) 1
Hint: (1)
(2)
ab
bc
ca
is
c) 2
d) 3
ac bc ab


 1
b
a
c
(a  b) 2  c (b  c) 2  a (c  a ) 2  b


ab  c
bc  a
ac  b
Take LCM ‘abc’ and simplify .
5.
Reduce to the lowest terms
a 2  b2  ab  b2
ab
ab  a 2
a 2  b 2 ba  b 

ab
aa  b 
Hint:
6.
The product of (a + b – c) and (a – b + c) is: [NTSE NOVEMBER 2008 - KAR]
a) a2 – b2 – c2 – 2bc
b) a2 – b2 – c2 + 2bc
c) a2 + b2 + c2 + 2bc
d) a2 + b2 + c2 – 2bc
7.
The expression x2 – y2 – z2 + 2yz + x + y – z has a factor: [NTSE MAY 2007 - NAT]
a) x – y + z + 1
b) – x + y + z
c) x + y – z + 1
d) x – y – z + 1
8.
If x + y + z = 0, then a factor of the expression (x + y)3 + (y + z)3 + (z + x)3 is:
[NTSE MAY 2007 - NAT]
a) 3(x + y)(y + z)(z + x)
b) –3xyz
c) (x + y – z)
d) (x – y + z)
Class - 7
Quadratic Equations
INTRODUCTION
I
i) General form of Quadratic Equations  ax2 + bx + c = 0
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CFAL
MATHEMATICS- MODULE 1
ii) The roots of Quadratic Equations are also called zeros of the polynomial. Every Quadratic
Equations can be written as product of two linear polynomials.
ax2 + bx + c = (x - ) (x - )
Where  and  are zeros of ax2 + bx + c
II Finding roots using Factor Method
i) Express the equation in the form of ax2 + bx + c = 0
ii) Factorise the LHS expression
iii) Equate each factor to zero and find the roots of the Quadratic Equations

Example:
2x2 + 2 = 5x
2x2 – 5x + 2 = 0
2x2 – 4x – x + 2 = 0
(2x – 1) = 0
(x – 2) = 0
x= 2
1
are the roots of 2x2 + 2 – 5x = 0
2
III Finding roots using formula
To find roots of ax2 + bx + c = 0
Multiply by 4a : 4a2x2 + 4abx + 4ac = 0
i.e., (2ax)2 + 2(2ax)b = – 4ac
Add b2 to both sides: (2ax)2 + 2(2ax)b + b2 = b2 – 4ac
i.e., (2ax + b)2 = b2 – 4ac
Taking the square roots, 2ax+b = 
i.e., 2ax = – b 
b 2  4ac or x =
b 2  4ac
 b  b 2  4ac
.
2a
Example: Find the roots of 3x2 – 5x – 7 = 0.
 Here a = 3, b = – 5, c = – 7.
 b  b2  4ac 5  25  84 5  109
x=
.


2a
6
6
The roots are
5  109 5  109
.
,
6
6
CLASS ASSIGNMENT - 7
1. (a + 2b)2 – 8ab is equal to : [NTSE NOVEMBER 2006 - KAR]
a) a2 + 4b2
b) a2 - 4b2
c) (a – 2b)2
2
Hint: expand like ( a+b)
d) a2 + 2b2
2. The product of (2x + 3) and (3x + 2) is equal to : [NTSE NOVEMBER 2006 - KAR]
a) 6x2 + 6
b) 6x2 + 13x + 6
c) 6 (x2 + x + 1)
Hint: x2(product of x coefficient) + [2x(2)+3x(3)]+addition of constants
Centre For Advanced Learning, Kuntikan Junction, Mangalore-04, Tel.2249220
d) 5x2 + 6x + 6
27
CFAL
MATHEMATICS- MODULE 1
3. The solution of the equation x  25  x 2  1 is
a) x = 1
b) x = -3
c) x = 4
there is mismatch of solution.
4. In the equation 2x + 1 + 4x = 80 the value of x is
1) less than 3
2) equal to 3
d) x =  5
3) greater than 3
4) none of these
there is mismatch of solution.
5. The equation x  5  x  5  6 has
1) onlyone root
2) exactly two roots
3) infinitely many roots
4) no root
there is mismatch of solution.
6. Solve the equation 10x2 – 3x – 1 = 0 using quadratic formula.
Hint: 10=5x2; -5x+3x=-3x
7. Solve the equation 3x2 + 2x + 16 = 0 using quadratic formula
Hint:
 b  b 2  4ac
X=
2a
a=3, b=2, c=16
If the equation x2 + 2 (k + 1)x + 9 = 0 has equal roots, then the value of k is,
a) 2
b) 3
c) 4
d) 5
Hint: b2-4ac=0; b=2(k+1);a=1;c=9 now solve for k
8.
9. If  and  are the roots of the equation ax2 + 2bx + c = 0, then
a)
2b
ac
b)
2b
ac
c)
 2b


is equal to,



b
d)
ac
ac

  





Hint ;
α + β = -b/a; α β =c/a from the equation substitute the values and calculate.
10. The equation
a) b,
a2
b
x a a b
   has two toots. They are
a x b a
b2
b) a, b
c) a ,
a
d)
a b
,
b a
Hint: substitute each option and check the answer.
HOME ASSIGNMENT – 7
1. Solve 2(x2 – 1) – 3(x2 - 2) = 0.
2. Solve the equation 3x -
5
= 2.
x
Hint; 3x2 – 2x – 5 = 0
3. Solve the equation 13  x 2 = x + 5.
Hint: square both sides and make a equation in the standard form. Factorize and find x.
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CFAL
MATHEMATICS- MODULE 1
2
4.
m2
+ m - 156 .
Check the problem
5. A quadratic equation with rational coefficients has both roots real and irrational, if the discriminant
a) is a perfect square
b) is positive, but not a perfect square
c) is negative , but not a perfect square d) zero
6. The equation 9x2 + 3kx + 4 = 0 has repeated roots, when k is equal to
a)  1
b)  2
c)  3
d) 4
Hint : b2 = 4ac = 0
7. The roots of the equation (q – r)x2+ (r – p)x + (p – q) = 0 are
a)
pq
,1
qr
b)
qr
,1
pq
c)
rp
,1
pq
d)
rp
,1
qr
hint; let 1 and k be the two roots of the equation.
1 k 
 (r  p)
qr
Now solve for k.
8. For the equation
a  x 2a  x 3a  x


, which of the following relations holds good?
a
2a
3a
a) 6x = 7a
9. The value of x in
b) ax = 0
c) 7x = 6a
d) a = x
x  bc x  ca x  ab


 a  b  c is
bc
ca
ab
a) a  b  c
b) abc
c) ab + bc + ca
d) a2 + b2 + c2
2
10. The equation whose roots are twice the roots of x – 3x + 3 = 0 is,
a) x2 – 6x + 12 = 0
b) x2 – 3x + 6 = 0
2
c) 2x – 3x + 3 = 0
d)4 x2 – 6x + 3 = 0
Hint: let the roots be 2a,2b. where ‘a’ and ‘b’ are roots of the given equation. Since a=x/2
x 2 3x

3 0
4
2
x 2  6 x  12  0
11. One of the value of p such that 3x2 – 2x + p = 0 and 6x2 – 17x + 12 = 0 have a common root
is,
a)
3
8
b)
8
3
c) 
8
3
d) 
3
8
Hint: find the roots of the 2nd equation. Substitute in place of x in equation 1. And solve for ‘p’
12. Ajay and Vijay solved an equation. In solving it, Ajay made a mistake in the constant term only
and got the roots as 8 and 2, while Vijay made a mistake in the coefficient of x only and obtained
the roots as -9 and -1. The correct roots of the equation are:
a) 8, -1
Hint:
b) -9, 2
b
 10
a
c) -8, -2
d) 9, 1
c
9
a
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Substitute b=-10a; c=9a in the standard equation. And solve for x.
13. Solve
x b a b
  
b x b a
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