Squares and Square Roots

Introduction

You know that the area of a square = side × side (where ‘side’ means ‘the length of a side’). Study the following table.

What is special about the numbers 4, 9, 25, 64 and other such numbers?

Since, 4 can be expressed as 2 × 2 = 2² , 9 can be expressed as 3 × 3 = 3² , all such numbers can be expressed as the product of the number with itself.

Such numbers like 1, 4, 9, 16, 25, ... are known as square numbers.

In general, if a natural number m can be expressed as n² , where n is also a natural number, then m is a square number.

Is 32 a square number?, We know that 5² = 25 and 6² = 36. If 32 is a square number, it must be the square of a natural number between 5 and 6. But there is no natural number between 5 and 6. Therefore 32 is not a square number.

Consider the following numbers and their squares.

From the above table, can we enlist the square numbers between 1 and 100? Are there any natural square numbers upto 100 left out? You will find that the rest of the numbers are not square numbers.

The numbers 1, 4, 9, 16 ... are square numbers. These numbers are also called perfect squares.

1.Properties of Square Numbers:

· Following table shows the squares of numbers from 1 to 20.

Study the square numbers in the above table. What are the ending digits (that is, digits in the units place) of the square numbers?

All these numbers end with 0, 1, 4, 5, 6 or 9 at units place. None of these end with 2, 3, 7 or 8 at unit’s place. Can we say that if a number ends in 0, 1, 4, 5, 6 or 9, then it must be a square number.

· Study the following table of some numbers and their squares and observe the one’s place in both.

The following square numbers end with digit 1.

Write the next two square numbers which end in 1 and their corresponding numbers. You will see that if a number has 1 or 9 in the units place, then it’s square ends in 1.

· Let us consider square numbers ending in 6.

We can see that when a square number ends in 6, the number whose square it is, will have either 4 or 6 in unit’s place.

Can you find more such rules by observing the numbers and their squares (Table 1)

· Consider the following numbers and their squares.

If a number contains 3 zeros at the end, how many zeros will its square have ? What do you notice about the number of zeros at the end of the number and the number of zeros at the end of its square.

2.Some More Interesting Patterns

1. Adding triangular numbers.

triangular numbers (numbers whose dot patterns can be arranged as triangles)

If we combine two consecutive triangular numbers, we get a square number, like

2. Numbers between square numbers

Let us now see if we can find some interesting pattern between two consecutive square numbers.

Between1² (=1) and 2² (= 4) there are two (i.e., 2 × 1) non square numbers 2, 3.

Between 2² (= 4) and 3² (= 9) there are four (i.e., 2 × 2) non square numbers 5, 6, 7, 8.

Now, 3² = 9, 4² = 16

Therefore, 4² – 3² = 16 – 9 = 7

Between 9(3² ) and 16( 4² ) the numbers are 10, 11, 12, 13, 14, 15 that is, six non-square numbers which is 1 less than the difference of two squares.

We have 4² = 16 and 5² = 25

Therefore, 5² – 4² = 9

Between 16(= 4² ) and 25(= 5² ) the numbers are 17, 18, ... , 24 that is, eight non square numbers which is 1 less than the difference of two squares.

Consider 7² and 6² . Can you say how many numbers are there between 6² and 7² ?

If we think of any natural number n and (n + 1), then,

(n + 1)² – n² = (n² + 2n + 1) – n² = 2n + 1.

We find that between n² and (n + 1)² there are 2n numbers which is 1 less than the difference of two squares.

Thus, in general we can say that there are 2n non perfect square numbers between the squares of the numbers n and (n + 1). Check for n = 5, n = 6 etc., and verify.

3. Adding odd numbers

Consider the following

1 [one odd number] = 1 = 1²

1 + 3 [sum of first two odd numbers] = 4 = 2²

1 + 3 + 5 [sum of first three odd numbers] = 9 = 3²

1 + 3 + 5 + 7 [... ] = 16 = 4²

1 + 3 + 5 + 7 + 9 [... ] = 25 = 5²

1 + 3 + 5 + 7 + 9 + 11 [... ] = 36 = 6²

So we can say that the sum of first n odd natural numbers is n²

Looking at it in a different way, we can say: ‘If the number is a square number, it has to be the sum of successive odd numbers starting from 1.

Consider the number 25. Successively subtract 1, 3, 5, 7, 9, ... from it

(i) 25 – 1 = 24 (ii) 24 – 3 = 21 (iii) 21 – 5 = 16 (iv) 16 – 7 = 9 (v) 9 – 9 = 0

This means, 25 = 1 + 3 + 5 + 7 + 9. Also, 25 is a perfect square.

Now consider another number 38, and again do as above.

(i) 38 – 1 = 37 (ii) 37 – 3 = 34 (iii) 34 – 5 = 29 (iv) 29 – 7 = 22 (v) 22 – 9 = 13 (vi) 13 – 11 = 2 (vii) 2 – 13 = – 11

This shows that we are not able to express 38 as the sum of consecutive odd numbers starting with 1. Also, 38 is not a perfect square.

So we can also say that if a natural number cannot be expressed as a sum of successive odd natural numbers starting with 1, then it is not a perfect square.

4. A sum of consecutive natural numbers

Consider the following

we can express the square of any odd number as the sum of two consecutive positive integers.

5. Product of two consecutive even or odd natural numbers

11 × 13 = 143 = 12² – 1

Also 11 × 13 = (12 – 1) × (12 + 1)

Therefore, 11 × 13 = (12 – 1) × (12 + 1) = 12² – 1

Similarly, 13 × 15 = (14 – 1) × (14 + 1) = 14² – 1

29 × 31 = (30 – 1) × (30 + 1) = 30² – 1

44 × 46 = (45 – 1) × (45 + 1) = 45² – 1

So in general we can say that (a + 1) × (a – 1) = a² – 1.

6. Some more patterns in square numbers

Observe the squares of numbers; 1, 11, 111 ... etc. They give a beautiful pattern:

1² = 1

11² = 1 2 1

111 ² = 1 2 3 2 1

1111² = 1 2 3 4 3 2 1

11111² = 1 2 3 4 5 4 3 2 1

11111111² = 1 2 3 4 5 6 7 8 7 6 5 4 3 2 1

Another interesting pattern.

7² = 49

67² = 4489

667² = 444889

6667² = 44448889

66667² = 4444488889

666667² = 444444888889

Example: What will be the unit digit of the squares of the following numbers?

(i) 81 (ii) 272 (iii) 799

Solution:

(i) 81

(81)² = 6561

unit digit of square of 81 is 1.

(ii) 272

(272)² = 73984

unit digit of square of 272 is 4.

(iii) 799

(799)² = 68401

unit digit of square of 799 is 1.

Example: Observe the following pattern and find the missing digits.

11² = 121

101² = 10201

1001² = 1002001

100001² = 1 ......... 2 ......... 1

10000001² = ...........................

Solution:

11² = 121

101² = 10201

1001² = 1002001

100001² = 10000200001

10000001² = 100000020000001

Example: Without adding, find the sum.

(i) 1 + 3 + 5 + 7 + 9

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 +19

(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23

Solution:

(i) 1 + 3 + 5 + 7 + 9

we know that the sum of first n odd natural numbers is n² .

here, n= total number of odd no’s

so, n= 5

Therefore, n² = 5² = 25

1 + 3 + 5 + 7 + 9 = 25

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 +19

we know that the sum of first n odd natural numbers is n² .

here, n= total number of odd no’s

so, n= 10

Therefore, n² = 10² = 100

(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23

we know that the sum of first n odd natural numbers is n² .

here, n= total number of odd no’s

so, n= 12

Therefore, n² = 12² = 144

3.Finding the Square of a Number

Squares of small numbers like 3, 4, 5, 6, 7, ... etc. are easy to find. But can we find the square of 23 so quickly?

The answer is not so easy and we may need to multiply 23 by 23.

There is a way to find this without having to multiply 23 × 23.

We know 23 = 20 + 3

Therefore 23² = (20 + 3)² = 20(20 + 3) + 3(20 + 3)

= 20² + 20 × 3 + 3 × 20 + 3²

= 400 + 60 + 60 + 9 = 529

Example :

Find the square of the following numbers without actual multiplication.

(i) 39 (ii) 42

Solution:

(i) 39² = (30 + 9)² = 30(30 + 9) + 9(30 + 9)

= 302 + 30 × 9 + 9 × 30 + 92

= 900 + 270 + 270 + 81 = 1521

(ii) 42² = (40 + 2)² = 40(40 + 2) + 2(40 + 2)

= 40² + 40 × 2 + 2 × 40 + 2²

= 1600 + 80 + 80 + 4 = 1764

3.1 Other patterns in squares

Consider the following pattern:

25² = 625 = (2 × 3) hundreds + 25

35² = 1225 = (3 × 4) hundreds + 25

75² = 5625 = (7 × 8) hundreds + 25

125² = 15625 = (12 × 13) hundreds + 25

Consider a number with unit digit 5, i.e., a5

(a5)² = (10a + 5)²

= 10a(10a + 5) + 5(10a + 5)

= 100a² + 50a + 50a + 25

= 100a(a + 1) + 25

= a(a + 1) hundred + 25

3.2 Pythagorean triplets

Consider the following

3² + 4² = 9 + 16 = 25 = 5²

The collection of numbers 3, 4 and 5 is known as Pythagorean triplet. 6, 8, 10 is also a Pythagorean triplet, since

6² + 8² = 36 + 64 = 100 = 10²

Again, observe that

5² + 12² = 25 + 144 = 169 = 13² . The numbers 5, 12, 13 form another such triplet

For any natural number m > 1, we have (2m)² + (m² – 1)² = (m² + 1)² . So, 2m, m2 – 1 and m2 + 1 forms a Pythagorean triplet.

Example 1:

Write a Pythagorean triplet whose smallest member is 8.

Solution:

We can get Pythagorean triplets by using general form 2m, m² – 1, m² + 1.

Let us first take m² – 1 = 8

So, m² = 8 + 1 = 9

which gives m = 3

Therefore, 2m = 6 and m² + 1 = 10

The triplet is thus 6, 8, 10. But 8 is not the smallest member of this

So, let us try 2m = 8

then m = 4

We get m² – 1 = 16 – 1 = 15

and m² + 1 = 16 + 1 = 17

The triplet is 8, 15, 17 with 8 as the smallest member.

Example 2:

Find a Pythagorean triplet in which one member is 12

Solution:

If we take m2 – 1 = 12

Then, m2 = 12 + 1 = 13

Then the value of m will not be an integer.

So, we try to take m2 + 1 = 12. Again m2 = 11 will not give an integer value for m.

So, let us take 2m = 12

then m = 6

Thus, m2 – 1 = 36 – 1 = 35

and m2 + 1 = 36 + 1 = 37

Therefore, the required triplet is 12, 35, 37.

Note : All Pythagorean triplets may not be obtained using this form. For example another triplet 5, 12, 13 also has 12 as a member.

4 Square Roots

Study the following situations.

(a) Area of a square is 144 cm² . What could be the side of the square?

We know that the area of a square = side²

If we assume the length of the side to be ‘a’, then 144 = a²

To find the length of side it is necessary to find a number whose square is 144.

(b) What is the length of a diagonal of a square of side 8 cm ?

Can we use Pythagoras theorem to solve this ?

Can we use Pythagoras theorem to solve this ?

We have, AB² + BC² = AC²

i.e., 8² + 8² = AC²

or 64 + 64 = AC²

or 128 = AC²

Again to get AC we need to think of a number whose square is 128.

In all the above cases, we need to find a number whose square is known. Finding the number with the known square is known as finding the square root.

4.1 Finding square roots

The inverse (opposite) operation of addition is subtraction and the inverse operation of multiplication is division. Similarly, finding the square root is the inverse operation of squaring.

We have, 1² = 1, therefore square root of 1 is 1

2² = 4, therefore square root of 4 is 2

3² = 9, therefore square root of 9 is 3

From the above, you may say that there are two integral square roots of a perfect square number. In this chapter, we shall take up only positive square root of a natural number. Positive square root of a number is denoted by the symbol .

For example: 4 = 2 (not –2); 9 = 3 (not –3) etc.

4.2 Finding square root through repeated subtraction

The sum of the first n odd natural numbers is n 2 ? That is, every square number can be expressed as a sum of successive odd natural numbers starting from 1.

Consider . Then,

(i) 81 – 1 = 80 (ii) 80 – 3 = 77 (iii) 77 – 5 = 72 (iv) 72 – 7 = 65

(v) 65 – 9 = 56 (vi) 56 – 11 = 45 (vii) 45 – 13 = 32 (viii) 32 – 15 = 17

(ix) 17 – 17 = 0

From 81 we have subtracted successive odd numbers starting from 1 and obtained 0 at 9th step.

Therefore = 9.

4.3 Finding square root through prime factorisation

Consider the prime factorisation of the following numbers and their squares.

How many times does 2 occur in the prime factorisation of 6? Once. How many times does 2 occur in the prime factorisation of 36? Twice. Similarly, observe the occurrence of 3 in 6 and 36 of 2 in 8 and 64 etc.

You will find that each prime factor in the prime factorisation of the square of a number, occurs twice the number of times it occurs in the prime factorisation of the number itself. Let us use this to find the square root of a given square number, say 324.

We know that the prime factorisation of 324 is 2 324

324 = 2 × 2 × 3 × 3 × 3 × 3 2 162

3 81

3 27

3 9

3

By pairing the prime factors,

we get 324 = 2 × 2 × 3 × 3 × 3 × 3 = 2² × 3² × 3² = (2 × 3 × 3)² 2 256

So, = 2 × 3 × 3 = 18 2 128

2 64

Similarly can you find the square root of 256? Prime factorisation of 256 is 2 32

256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 2 16

By pairing the prime factors we get, 2 8

256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = (2 × 2 × 2 × 2)² 2 4

Therefore, = 2 × 2 × 2 × 2 = 16 2

Is 48 a perfect square?

We know 48 = 2 × 2 × 2 × 2 × 3

Since all the factors are not in pairs so 48 is not a perfect square

Example 1: 2 6400

Find the square root of 6400 2 3200

2 1600

Solution: 2 800

Write 6400 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 2 400

Therefore = 2 × 2 × 2 × 2 × 5 = 80 2 200

2 100

2 50

5 25

5

Example 2:

Is 90 a perfect square? 2 90

Solution: 3 45

We have 90 = 2 × 3 × 3 × 5 3 15

The prime factors 2 and 5 do not occur in pairs. 5

Therefore, 90 is not a perfect square.

That 90 is not a perfect square can also be seen

from the fact that it has only one zero.

Example 3:

Find the smallest number by which 9408 must be divided so that the quotient is a perfect square. Find the square root of the quotient.

Solution:

We have, 9408 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 7 × 7

If we divide 9408 by the factor 3, then

9408 ÷ 3 = 3136 = 2 × 2 × 2 × 2 × 2 × 2 × 7 × 7 which is a perfect square.

Therefore, the required smallest number is 3.

And, = 2 × 2 × 2 × 7 = 56.

4.4 Finding square root by division method

When the numbers are large, even the method of finding square root by prime factorisation becomes lengthy and difficult. To overcome this problem we use Long Division Method.

The smallest 3-digit perfect square number is 100 which is the square of 10 and the greatest 3-digit perfect square number is 961 which is the square of 31. The smallest 4-digit square number is 1024 which is the square of 32 and the greatest 4-digit number is 9801 which is the square of 99.

The use of the number of digits in square root of a number is useful in the following method:

· Consider the following steps to find the square root of 529.

Can you estimate the number of digits in the square root of this number?

Step 1 Place a bar over every pair of digits starting from the digit at one’s place. If the number of digits in it is odd, then the left-most single digit too will have a bar.

Thus we have, .

2

Step 2 Find the largest number whose square is less than or equal to the number 2

under the extreme left bar (2² < 5 < 3² ). Take this number as the divisor -4

and the quotient with the number under the extreme left bar as the 1

dividend (here 5). Divide and get the remainder (1 in this case).

2

Step 3 Bring down the number under the next bar (i.e., 29 in this case) to the 2

right of the remainder. So the new dividend is 129. -4

1 29

Step 4 Double the quotient and enter it with a blank on its right.

Step 5 Guess a largest possible digit to fill the blank which will also become 2

the new digit in the quotient, such that when the new divisor is multiplied 2

to the new quotient the product is less than or equal to the dividend. -4

In this case , 42 × 2 = 84. 4_ 1 29

As 43 × 3 = 129 so we choose the new digit as 3. Get the remainder.

Step 6 Since the remainder is 0 and no digits are left in the given number, 23

therefore, = 23. 2

-4

43 1 29

-1 29

0

· Now consider

Step 1 Place a bar over every pair of digits starting from the one’s digit. ( ).

Step 2 Find the largest number whose square is less than or equal to the number under the left- most bar (6² < 40 < 7² ). Take this number as the divisor and the number under the left- most bar as the dividend. Divide and get the remainder i.e., 4 in this case.

Step 3 Bring down the number under the next bar (i.e., 96) to the right of the remainder. The new dividend is 496.

Step 4 Double the quotient and enter it with a blank on its right.

Step 5 Guess a largest possible digit to fill the blank which also becomes the new digit in the quotient such that when the new digit is multiplied to the new quotient the product is less than or equal to the dividend. In this case we see that 124 × 4 = 496. So the new digit in the quotient is 4. Get the remainder.

Step 6 Since the remainder is 0 and no bar left, therefore,

Estimating the number

We use bars to find the number of digits in the square root of a perfect square number.

= 23. And

In both the numbers 529 and 4096 there are two bars and the number of digits in their square root is 2. Can you tell the number of digits in the square root of 14400?

By placing bars we get . Since there are 3 bars, the square root will be of 3 digit.

Example :

Find the square root of : (i) 729 (ii) 1296

Solution:

5 Square Roots of Decimals

· Consider

Step 1 To find the square root of a decimal number we put bars on the integral part (i.e., 17) of the number in the usual manner. And place bars on the decimal part(i.e., 64) on every pair of digits beginning with the first decimal place. Proceed as usual. We get .

Step 2 Now proceed in a similar manner. The left most bar is on 17 and 4² < 17 < 5² . Take this number as the divisor and the number under the left-most bar as the dividend, i.e., 17. Divide and get the remainder.

Step 3 The remainder is 1. Write the number under the next bar (i.e., 64) to the right of this remainder, to get 164.

Step 4 Double the divisor and enter it with a blank on its right. Since 64 is the decimal part so put a decimal point in the quotient.

Step 5 We know 82 × 2 = 164, therefore, the new digit is 2. Divide and get the remainder.

Step 6 Since the remainder is 0 and no bar left, therefore

Example : Find the square root of 12.25.

Solution:

Example : Area of a square plot is 2304 m² . Find the side of the square plot.

Solution:

Area of square plot = 2304 m²

Therefore, side of the square plot = m

We find that, = 48

Thus, the side of the square plot is 48 m.

Example : There are 2401 students in a school. P.T. teacher wants them to stand in rows and columns such that the number of rows is equal to the number of columns. Find the number of rows.

Solution:

Let the number of rows be x

So, the number of columns = x

Therefore, number of students = x × x = x²

Thus, x² = 2401 gives x = = 49

The number of rows = 49.

6 Estimating Square Root

· Consider the following situations:

1. Deveshi has a square piece of cloth of area 125 cm² . She wants to know whether she can make a handkerchief of side 15 cm. If that is not possible she wants to know what is the maximum length of the side of a handkerchief that can be made from this piece.

2. Meena and Shobha played a game. One told a number and other gave its square root. Meena started first. She said 25 and Shobha answered quickly as 5. Then Shobha said 81 and Meena answered 9. It went on, till at one point Meena gave the number 250. And Shobha could not answer. Then Meena asked Shobha if she could atleast tell a number whose square is closer to 250.

In all such cases we need to estimate the square root.

We know that 100 < 250 < 400 and = 10 and = 20.

So 10 < < 20

But still we are not very close to the square number.

We know that 15² = 225 and 16² = 256

Therefore, 15 < < 16 and 256 is much closer to 250 than 225.

So, is approximately 16.