Question

How do you get rid of the square root sign in $\sqrt {147{m^3}{n^3}} $ ?

Answer 1

Hint: For this we have to calculate the prime factor of $147{m^3}{n^3}$. Prime factorization: It’s the process where the original given number is expressed as the product of prime numbers. Then we have to make the group of the same prime number.

Complete step by step solution:
Given
$\sqrt {147{m^3}{n^3}} $
We have to get rid of the square root sign in $\sqrt {147{m^3}{n^3}} $ ,
Now we know we have to do prime factorization of $147{m^3}{n^3}$ .
Prime factorization: It’s the process where the original given number is expressed as the product of prime numbers. Prime factors are those numbers or factors which are greater than one and have exactly two factors one is the number itself and other is one .
Prime Factorization of $147{m^3}{n^3}$ is ,
$147{m^3}{n^3} = 3 \times 7 \times 7 \times m \times m \times m \times n \times n \times n$
$ = 3 \times {7^2} \times {m^2} \times m \times {n^2} \times n$
Now,
$\sqrt {147{m^3}{n^3}} = \sqrt {3 \times {7^2} \times {m^2} \times m \times {n^2} \times n} $
$ = 7mn\sqrt {3mn} $
We know that $\sqrt a = {(a)^{\dfrac{1}{2}}}$ ,

Therefore , we will get , $ = 7mn{(3mn)^{\dfrac{1}{2}}}$

Additional Information: Prime factorization is a way to write a composite number as the product of prime factors. Prime factors are those numbers or factors which are greater than one and have exactly two factors one is the number itself and other is one . There are basically two ways to find prime factorization namely ,
a) by division method
b) by factor tree

Note: We can also write this $\sqrt {147{m^3}{n^3}} $ as ${(147)^{\dfrac{1}{2}}}{(m)^{\dfrac{3}{2}}}{(n)^{\dfrac{3}{2}}}$ since $\sqrt a = {(a)^{\dfrac{1}{2}}}$ . If we have questions with similar nature as that of above can be approached in a similar manner and we can solve it easily and easily get rid of the square root.