Squares & Square Roots

We have already studied that the square of a number is the product of the number with the number itself. For example a given number is x, the square of x is (x x x), denoted by x ² .

e.g. (9) ² = 9 x 9 = 81

(5) ² = 5x 5 = 25

(10) ² = 10x 10 = 100

(1.2) ² = 1.2x 1.2 = 1.44

Perfect Squares or Square Numbers

A natural number is called a perfect square or a square number if it is the square of some natural numbers. It is always expressible as the product of equal factors.

e.g.

49 = 7x 7 = 7 ²

144 =2 x 2 x 2 x 2 x 3 x 3 = 12x 12 = 12 ²

225 = 5x 5x 3 x 3 = 15x 15 = 15 ²

1000= 2 x 2 x 5x 5 = 10x 10 = 10 ²

Example. Show that 500 is not a perfect square.

Ans. Resolving 500 in to prime factors, we get

500 = 2 x 2 x 5x 5 x 5

Making pairs of equal factors, we find that 5 is unpaired

Hence 500 is not a perfect square.

Properties of Perfect Squares

Example 1. A number ending in 2, 3, 7 or 8 is never a perfect square

e.g. 62,103,113,148 ,237,107.

Example 2. A number ending in an odd number of zeros is never a perfect square.

e.g. 10,3000 , 13000, 120, 100000.

Example 3. The square of a even number is always even.

e.g. 12 ² = 144, 16 ² = 256 30 ² = 900 etc.

Example 4. The square of an odd number is always odd.

E.g. 5 ² = 25, 7 ² = 49, 13 ² =169, 11 ² = 121 etc.

Example 5. The square of a proper fraction is smaller than the fraction.

e.g. and since 9x 5 < 25x 3.

Example 6. For every natural number n we have

(n + 1) ² – n ² = (n + 1 + n) (n + 1 – n) = {(n + 1) + n}

{(25) ² – (24) ² } = (25 + 24) = 49

{(70) ² – (69) ² } = (70 + 69) = 139

Example 7. For every natural number n, we have sum of the first n odd natural numbers = n ²

E.g. {1+3+5+7+9+11+13}=Sum of first 6 odd numbers=6 ² =36

{1+3+5+7+9+11+13+15+17+19}}=Sum of first 10 odd numbers=10 ² =100

Pythagorus Triplets

A triplet (m, n, p) of three natural number. m, n and p is called a Pythagoras triplet of m ² + n ² = p ²

E.g. (3, 4, 5) (5, 12, 13), (8, 15, 17), etc.

Example 8. For every natural number m greater than 1, (2m, m ² – 1, m ² + 1) is a Pythagoras triplet.

Find the Pythagoras triplet where smallest number is 10.

Ans. For every natural number m > 1, (2m, m ² – 1, m ² + 1) is a Pythagoras triplet

For every natural number m > 1, (2m, m ² – 1, m ² + 1) is a Pythagoras triplet.

Put 2m = 12 m = 6

m ² – 1 = 35

m ² + 1 = 36

triplet (12, 35, 37)

Some Interesting Patterns of Square Numbers

Sum of first two is 4

Sum of second and third is 9

Sum of third and forth is 16… and so on

Thus n ^th triangular number is given by

Part I. There are 2n non-perfect square numbers between two consecutive square number n ² and (n + 1) ² .

e.g. Between the number 5 ² and4 ² .Total 8 numbers are between these two numbers.

Between the number 50 ² and49 ² .Total 98 numbers are between these two numbers.

Part II. The sum of first n odd natural numbers in n ²

n = 1 1 = 1 = 1 ²

n = 2 1 + 3 = 4 = 2 ²

n = 3 1 + 3 + 5 = 9 = 3 ²

n = 4 1 + 3 + 5 + 7 = 16 = 4 ²

n = 5 1 + 3 + 5 + 7+9 = 25 = 5 ²

n=10 1+3+5+7+9+11+13+15+17+19 =10 ²

Part III. If 1 is added to the product of two consecutive odd natural numbers, it is equal to the square of the only even natural number between them.

In general

(2n – 1) x (2n + 1) + 1 = 4n ² = (2n) ²

this pattern is very useful to find the result

1 x 3 + 1 = 4 = 2 ²

3 x 5 + 1 = 16 = 4 ²

5 x 7 + 1 = 36 = 6 ²

7 x 9 + 1 = 64 = 8 ² etc.

In reverse way we can factorize (2n) ² -1 in two factors as (2n – 1) x (2n + 1)

Part I V . If 1 is added to the product of two consecutive even natural numbers, it is equal to the square of the only odd natural number between them.

In general

2n (2n + 2) + 1 = 4n ² + 4n + 1 = (2n + 1) ²

E.g

2 x 4 + 1 = 9 = 3 ²

4 x 6 + 1 = 25 = 5 ²

6 x 8 + 1 = 49 = 7 ²

8 x 10 + 1 = 81 = 9 ²

10x 12 + 1 = 121 = 11 ²

12x 14 + 1 = 169 = 13 ²

and so on ………………

In reverse way we can factorize (2n + 1) ² -1 in two factors as 2n (2n + 2)= 4n(n+1)

Part V. Squares of natural numbers having all digits 1 follow the following pattern.

11 ² = 121

111 ² = 12321

1111 ² = 1234321

11111 ² = 123454321

111111 ² = 12345654321

and so on …………

we can use this in many ways as

22 ² = 4 x121=484

333 ² = 9 x12321 =110889

222222 ² = 4 x12345654321= 49382737284

Using suitable patterns, compute the following

(i) (ii)

Ans. We have the following pattern

11 ² = 121

111 ² = 12321

1111 ² = 1234321

11111 ² = 123454321

111111 ² = 12345654321

(i)

= 81.

(ii)

= 49.

Part VI. Observe the following pattern

7 ² = 49

67 ² = 4489

667 ² = 444889

6667 ² = 4444889

66667 ² = 44444889

and so on

Some Short–Cuts to Find Squares

Three column method – This method is based upon an old Indian method of multiplying two numbers. It is convenient for finding squares of two digit numbers only. This method uses the identity (a + b) ² = a ² + 2ab + b ²

Step I : Make three columns and write the value of a ² , 2 x a x b and b ² respectively in these columns as follows:

As an illustration let us take ab = 34 = 3x10+4

a = 3 and b = 4

Step II : Undertake the units digit of b ² ( in column III) and add its tens digit, if any, to 2 x a x b (n column II)

Step III : Undertake the units digit in column II and add the number forward by tens and other digit, if any to a ² in column II.

Step IV : Under the number in column I

Step V: Write the understual digits at the bottom of each column to obtain the square of the given number.

In this case, we have 34 ² = 1156.

Some Particular Methods

Rule1: The square of a number of the form a 5 (where a is tens digit and 5 is units digit) is the number which each in 25 and has the number a(a + 1) before 25.

Example: Find the squares of the following numbers

1. 45
2. 95
3. 85

Ans. 1. Here we have

a = 4=> a(a + 1) = 4× 5 = 20

(45) ² = 2025

2. Here, we have

a = 9 => a(a + 1) = 9 x 10 = 90

\95 ² = 9025.

3. Here, we have

a = 8=> a(a + 1) = 8x 9 = 72

75 ² = 7225.

Rule 2 : The square of a number of the form ( 5x 10 + a ) (where a is units digit and 5 is ten digit) is equal to

(25 + a) x 100 + a ²

(5 a) ² = (25 + a) x 100 + a ²

Example : Find the squares of the following numbers

1. 51
2. 53
3. 56

Ans. 1. Here, we have

a = 1

(51) ² = (25 + 1) x 100 + 1 ² = 2600 + 01 = 2601.

2. Here, we have

a = 3

(53) ² = (25 + 3) x 100 + 3 ² = 2800 + 09

= 2809

3. Here, we have

a = 6

(56) ² = (25 + 6) x 100 + 6 ² = 3100 + 36 = 3136.

Rule 3 : The square of a three digit number (5x 100+ ax10 +b) (where a,b are integers from 0 to 9.

(5x 100+ ax10 +b) ² = (250 + ab) x 1000 + (ax10 + b) ²

Example: Find the squares of the following numbers

1. 516
2. 513
3. 504

Ans. 1. Given number is 516

a = 1, b= 6

(516) ² = (250 + 16) x 1000 + (16) ²

= 266000 + 256

= 266256.

2. Here, we have

a = 1, b = 3

(513) ² = (250 + 13) x 1000 + (13) ²

= 263000 + 169

= 263169.

3. Here, we have

a = 1, b = 3

(504) ² = (250 + 04) x 1000 + (04) ²

= 254000 + 16

= 254016.

Rule 4 : The square of a number abc…..5 (i.e., a number having 5 at units place is obtained by affixing 25 to we right of the number n(n + 1), where n = abc….

Example: Find the square of the following numbers

1. 225
2. 265
3. 1235

Ans. 1. 225

Here n = 22

n(n + 1) = 22 x 23

= 506

Hence 225 ² = 50625

2. 265

Here n = 26

n(n + 1) =26x 27

= 702

Hence 225 ² = 70225

3. 1235

Here n = 123

(n + 1) = 123 x 124

= 15252

Hence 1235 ² = 1525225

Related Pages

1. Class 7 Maths Worksheet
2. Class 7 Maths Notes