# Squares and Square Roots - Class 8 : Notes

Square number:
If any natural number m can be expressed as n2, where n is also a natural number, then m is known as a square number.
The square numbers are also called as perfect squares.

Example
:
Let m = 36.
Now, 36 can be expressed as 62, where 6 is a natural number. Therefore, 36 is a square number.

Square numbers of some natural numbers:

 Number Square Number Square 1 1 × 1 = 1 11 11 × 11 = 121 2 2 × 2 = 4 12 12 × 12 = 144 3 3 × 3 = 9 13 13 × 13 =169 4 4 × 4 = 16 14 14 ×1 4 = 196 5 5 × 5 = 25 15 15 × 15 = 225 6 6 × 6 = 36 16 16 × 16 = 256 7 7 × 7 = 49 17 17 × 17 = 289 8 8 × 8 = 64 18 18 × 18 = 324 9 9 × 9 = 81 19 19 ×1 9 = 361 10 10 × 10 = 100 20 20 × 20 = 400

Properties of Square Numbers:
1. The unit’s place of square numbers can be 0, 1, 4, 5, 6 or 9.
No square number can end with 2, 3, 7 or 8.

2. If a number have 1 or 9 in its unit’s place, then square of that number will end with 1.
Example:

 Number Square 1 1 9 81 11 121 19 361 21 441 … …

3. If a number have 4 or 6 in its unit’s place, then square of that number will end with 6.
Example:

 Number Square 4 16 16 36 14 196 16 256 24 576 … …
4.There will always be even number of zeros at end of any square number.
Example:

 Number Square 10 100 20 400 80 6400 700 490000 900 810000 … …

5. On combining two consecutive triangular numbers we get a square number.
Example: 6. There are 2n non-perfect square numbers between the squares of the numbers n and (n+1).
Example: Between 32 = 9 and 42 = 16, there lies 6 numbers which are 10, 11, 12, 13, 14, and 15.

7. If the number is a square number, then it has to be the sum of successive odd numbers starting from 1.
Example: For 32 = 9, the sum of successive odd numbers from 1 will be 1+3+5 = 9.
Note: 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.

8. Square number can be summation of two consecutive natural numbers.
Example: 52 = 25 = 12 + 13; 72 = 49 = 24 + 25, etc.

9. Product of two consecutive even or odd natural numbers.
Example: 11 × 13 = (12-1) × (12+1) = 122 – 1;          13 × 15 = (14-1) × (14+1) = 142 – 1
So, in general (a+1) × (a-1) = a2 – 1.

10. Some interesting patterns in square numbers
(i)
12 =                                                      1
112 =                                        1          2          1
1112 =                          1          2          3          2          1
11112 =            1          2          3          4          3          2          1

(ii)
72 = 49
672 = 4489
6672 = 444889
66672 = 44448889

Some Examples based on Square Numbers:
Example 1: Find the unit digit of the square of the following numbers:
(a)           26837                  (b)           456
SolutionWe know that, the unit digit of any square number will end with the square of the unit digit of the given number.
(a) The unit digit of 26837 is 7. Now, square of 7 is 49.
Hence, the unit digit of 26837 will be 9.
(b) The unit digit of 456 is 6. Now, square of 6 is 36.
Hence, the unit digit of 456 will be 6.

Example 2: Find the quantity of number which can lie between square of 34 and 35.
SolutionWe know that, 2n numbers lie between n and (n+1).
Hence, between square of 34 and 35 there will 2 x 34 = 68 numbers.

Example 3: Find without adding the sum of 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23.
SolutionWe know that, the sum of first n odd numbers is n2.
Hence, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = 122 = 144.

Finding the Square of a Number:
1. A number can be divided into two parts, such that the square of those numbers are known.
Thus, x2 = (a +b)2, where (a + b) = x and values of square of a & b are known.

Example: 252 = (20 + 5) 2
= 202 + 20 x 5 + 5 x 20 + 52
= 400 + 100 +100 + 25 = 625.

2. For numbers ending with 5, follow (a5) 2 = a x (a+1) x 100 + 25
Example: 352 = 3 x (3+1) x 100 +25 = 1225

3. Pythagorean triplets
If sum of two square numbers results into a square number, then all these three numbers form a Pythagorean triplet.
Example 1: 32 + 42 = 9 + 16 = 25 = 52, so 3, 4 and 5 is known as Pythagorean Triplet.
In general, 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 2: Write a Pythagorean triplet having 18 as one its member.
Solution: Let’s take m2 + 1 = 18 i.e. m2  = 17. Thus, value of m will not be an integer.
Taking m2 - 1 = 18 i.e. m2  = 19. Thus, again value of m will not be an integer.
Now, let 2m = 18 i.e. m = 9.
Hence, m2 – 1 = 81 – 1 = 80 and m2 + 1 = 81 + 1 = 82.
Thus, the Pythagorean triplet is 18, 80 and 82.

Finding the Square Root of a Number:
Square root is the inverse operation of squaring. The positive square root of a number is denoted by the symbol √.

Example: √9 = 3. It cannot be -3.

1. Repeated subtraction:
In this method, given square number is subtracted from successive odd natural numbers starting from 1 until result of subtraction does not become 0.

Example 1: Find √16.
Solution:
(1) 16 – 1 =15               (2) 15 – 3 = 12              (3) 12 – 5 = 7                (4) 7 – 7 = 0
We can see the result is zero at the fourth step. Thus, √16 = 4.

Example 2: Find square root of 169 using repeated subtraction.
Solution:
(1) 169 – 1 =168           (2) 168 – 3 = 165          (3) 165 – 5 = 160          (4) 160 – 7 = 153
(5) 153 – 9 =144           (6) 144 – 11 = 133        (7) 133 – 13 = 120        (8) 120 – 15 = 105
(9) 105 – 17 =88           (10) 88 – 19 = 69          (11) 69 – 21 = 48          (12) 48 – 23 = 25
(13) 25 – 25 =0
We can see the result is zero at the thirteenth step. Thus, √169 = 13.

2. Prime Factorization
Let us understand this method by an example.

Example: Find √324 using prime factorization.
Solution: The prime factors of 324 = 2 x 2 x 3 x 3 x 3 x 3.

Note: The prime factors of any square number exist in pair.
Now, to find square root from these prime factors, re-write a number once per pair i.e. 2 x 3 x 3 as for this example.
The answer of (2 x 3 x 3 = 18) will be the result of √324.
Thus, √324 = 18.

3. Division Method
Follow the steps given below to understand this method:
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.
Step 2: Find the largest number whose square is less than or equal to the number under the extreme left bar. Take this number as the divisor and the quotient with the number under the extreme left bar as the dividend. Divide and get the remainder.
Step 3: Bring down the number under the next bar to the right of the remainder.
Step 4: Double the divisor and enter it with a blank on its right.
Step 5: Guess a largest possible digit to fill the blank which will also become the new digit in the quotient, such that when the new divisor is multiplied to the new quotient the product is less than or equal to the dividend.
Step 6: Repeat steps 3, 4 and 5 until remainder does not become 0.

Example 1: Find √729 using division method.
Solution: Thus, √729 = 27.

Example 2: Find √1024 using division method.
Solution: Thus, √1024 = 32.

Finding Square Roots of Decimals:
Follow the steps given below to understand this method:
Step 1: To find the square root of a decimal number we put bars on the integral part of the number in the usual manner. And place bars on the decimal part on every pair of digits beginning with the first decimal place. Proceed as done in the above method.
Step 2:  Find the largest number whose square is less than or equal to the number under the extreme left bar. Take this number as the divisor and the number under the left-most bar as the dividend. Divide and get the remainder.
Step 3: Write the number under the next bar to the right of this remainder.
Step 4: Double the divisor and enter it with a blank on its right.
Step 5: Guess a largest possible digit to fill the blank which will also become the new digit in the quotient, such that when the new divisor is multiplied to the new quotient the product is less than or equal to the dividend. Divide and get the remainder.
Step 6: Repeat steps 3, 4 and 5 until remainder does not become 0.

Example 1: Find √7.29 using division method.
Solution: Thus, √7.29 = 2.7

Example 2: Find √31.36 using division method.
Solution: Thus, √31.36 = 5.6

Estimating Square Root:
Approximation is the only way to find the estimated square root of a number.

Example: To find the estimate square root of 250.
Solution: We know that, 250 lies between 100 and 400 i.e. 100 < 250 < 400.
Now, √100 = 10 and √400 = 20. So, 10 < 250 < 20.
We can further approximate the numbers as we know that 152 = 225 and 162 = 256.
Thus, we can approximate square root of 250 as 16 as it is more nearer.

Some Examples based on Square Root of Numbers:

Example 1: Find the possible unit digit of following sqaure numbers:
(a)      998001            (b)       15129
Solution:
(a) We know that, if any square number ends with 1 then the unit digit of its square root can be 1 or 9.
(b) We know that, if any square number ends with 9 then the unit digit of its square root can be 3 or 7.

Example 2: Find the smallest number by which 768 be multiplied to obtain a perfect square number. Also, find the square root of that number.
SolutionThe prime factors of 768 are 2, 2, 2, 2, 2, 2, 2, 2, 3.
On forming groups, we have 768 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3.
We can see that, pair of 3 is missing. Hence, on multiplying 768 by 3 we will get a perfect square number.
Thus, 768 x 3 = 2304 is a perfect square.
Now, √2304 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 = 2 x 2 x 2 x 2 x 3 = 48.

Example 3: Find the smallest number by which 2800 be divided to obtain a perfect square number. Also, find the square root of that number.
SolutionThe prime factors of 2800 are 2, 2, 2, 2, 5, 5, 7.
On forming groups, we have 2800 = 2 x 2 x 2 x 2 x 5 x 5 x 7.
We can see that, pair of 7 is missing. Hence, on dividing 2800 by 7 we will get a perfect square number.
Thus, 2800 / 7 = 400 is a perfect square.
Now, √400 = 2 x 2 x 2 x 2 x 5 x 5 = 2 x 2 x 5 = 20.

Example 4: Find the smallest square number which is divisible by each of the numbers 8, 15 and 20.
SolutionThe LCM of 8, 15 and 20 = 2 x 2 x 2 x 3 x 5 =120.
Now, 120 is not a perfect square as 2, 3 and 5 have missing pairs.
Hence, 120 must be multiplied by 2, 3 and 5 i.e. 30 to get the perfect square.
Thus, the required square number = 120 x 30 = 3600.

Example 5: Find the smallest number which must be subtracted from 3250 to obtain a perfect square number. Also, find the square root of that number.
SolutionUsing division method, we have, Here, remainder is 1.
Thus, we must subtract 1 from 3250 to get perfect square number.
Therefore, required square number = 3250 – 1 = 3249.
And, √3249 = 57.

Example 6: Find the smallest number which must be added to 1825 to obtain a perfect square number. Also, find the square root of that number.
SolutionUsing division method, we have, Here, remainder is 61 which means square of 42 is less than 1825.
The next number after 42 is 43 whose square is 1849.
So, number to be added to 1825 = 432 – 1826 = 1849 - 1826 = 24.
Thus, we must add 24 to 1825 to get perfect square number.
Therefore, required square number = 1825 + 24= 1849.
And, √1849 = 43.

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