Question

Using prime factorization find the square root of $ 4761 $

Answer 1

Hint: Here we will use the different properties of the squares and square-roots considering that the square and square root cancel each other. That is $ n = \sqrt n \times \sqrt n $ . Here we will find the prime factors of the given number.

Complete step-by-step answer:
Prime factorization is defined as 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,}}...... $ $ 2 $ is the prime number as it can have only a $ 1 $ factor.
To get the prime factors of the given number, start dividing it with the least prime number and move to the greater number if not divisible by it.
 $ \sqrt {4761} = \sqrt {3 \times 3 \times 23 \times 23} $
The above equation can be re-written as,
 $ \sqrt {4761} = \sqrt {{3^2} \times {{23}^2}} $
When powers are equal, the bases can be written as the product of the terms taking the whole square.
 $ \sqrt {4761} = \sqrt {{{(3 \times 23)}^2}} $
Square and square-root cancel each other on the right-hand side of the equation.
 $ \sqrt {4761} = 3 \times 23 $
Find the product of the terms in the above expression –
 $ \sqrt {4761} = 69 $
This is the required solution.
So, the correct answer is “69”.

Note: Perfect square number is defined as the square of an integer, simply it is the product of the same integer with itself. For example - $ 25{\text{ = 5 }} \times {\text{ 5, 25 = }}{{\text{5}}^2} $ , normally it is denoted by n to the power two i.e. $ {n^2} $ . Square is the product of same number twice such as $ {n^2} = n \times n $ for Example square of $ 2 $ is $ {2^2} = 2 \times 2 $ shortened form of squared number is $ {2^2} = 2 \times 2 = 4 $ and square-root is represented by $ \sqrt {{n^2}} = \sqrt {n \times n} $ For Example: $ \sqrt {{2^2}} = \sqrt 4 = 2 $