if a natural number m can be expressed as n2, where n is also a natural number, then m is a square number.
1. Identity method
We know that (a+b)^{2} =a^{2} +2ab+b^{2}
Example: 23^{2}= (20+3)^{2} =400+9+120=529
2. Special Cases
(a5)^{2}
= a(a + 1) hundred + 25
Example: 25^{2}=2(3) hundred + 25 = 625
For any natural number m > 1, we have (2m)^{2} + (m^{2} – 1)^{2} = (m^{2} + 1)^{2} So, 2m, m^{2} – 1 and m^{2} + 1 forms a Pythagorean triplet
Example: 6,8,10
6^{2} +8^{2} =10^{2}
Square root of a number is the number whose square is given number
So we know that
m=n^{2} Square root of m
√m =n
Square root is denoted by expression √
1. Finding square root through repeated subtraction
We know sum of the first n odd natural numbers is n^{2}. So in this method we subtract the odd number starting from 1 until we get the reminder as zero. The count of odd number will be the square root
Consider 36 Then,
(i) 36 – 1 = 35
(ii) 35 – 3 = 32
(iii) 32 – 5 = 27
(iv) 27 – 7 = 20
(v) 20 – 9 = 11
(vi)11 – 11 = 0
So 6 odd number, Square root is 6
2. Finding square root through prime Factorisation
This method, we find the prime factorization of the number. We will get same prime number occurring in pair for perfect square number. Square root will be given by multiplication of prime factor occurring in pair
Consider
81
81=(3×3)×(3×3)
√81= 3×3=9
Finding square root by division method
This can be well explained with the example
79 videos408 docs31 tests


Explore Courses for Class 8 exam
