How do you find the square-root of $484?$

Hint: First of all we will find the prime factors of the given number and then will make a pair of factors and then will accordingly find the square-root. Square-root of number can be given as $\sqrt {n \times n} = \sqrt {{n^2}} = n$

Complete step-by-step answer:
Find the prime factors of the given number.
Prime factorization is the process of finding which prime numbers can be multiplied together to make the original number, where prime numbers are the numbers greater than $1$ and which are not the product of any two smaller natural numbers. For Example: $2,{\text{ 3, 5, 7,}}......$
Start dividing the given number with the least prime number $2$ , if it is further not divided then start dividing with the next least prime number that is $3$ and so on...
So, the factors of the given number can be expressed as –
$484 = 2 \times 2 \times 11 \times 11$
Take square root on both the sides of the equation-
$ \Rightarrow \sqrt {484} = \sqrt {2 \times 2 \times 11 \times 11} $
When the same number is multiplied twice, then it can be written as the square of that number.
$ \Rightarrow \sqrt {484} = \sqrt {{2^2} \times {{11}^2}} $
By using the law of power and exponent, when the powers are the same then the bases are written as its product.
$ \Rightarrow \sqrt {484} = \sqrt {{{(2 \times 11)}^2}} $
Square and square-root cancel each other on the right hand side of the equation.
$ \Rightarrow \sqrt {484} = 2 \times 11$
Find the product on the right hand side of the equation.
$ \Rightarrow \sqrt {484} = 22$

Note: Remember that the factors of the given number can be calculated by using the long division method or by factor tree method. To find the factors you should convert the given number into prime factors and factors should be in pairs of two. The perfect squares always have factors in pairs of two same numbers.