
How do you simplify $ \sqrt {\dfrac{{21}}{4}} $ ?
Answer
515.1k+ views
Hint: Here the symbol or the radical suggests us the square-root of the terms. Here we will find the prime factors of the given terms and then accordingly take out its square-root and then will simplify for the required resultant value.
Complete step by step solution:
Take the given expression: $ \sqrt {\dfrac{{21}}{4}} $
Find the prime factors of the terms in the above expression. Factors of any term are terms which when multiplied together then it gives us the original term.
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 defined as the numbers which are greater than $ 1 $ and which are also not the product of any two smaller natural numbers.
$ \sqrt {\dfrac{{21}}{4}} = \sqrt {\dfrac{{3 \times 7}}{{2 \times 2}}} $
When the term in division is whole square-rooted we can rewrite by applying the square-root of the numerator upon the square of the denominator part.
$ \sqrt {\dfrac{{21}}{4}} = \dfrac{{\sqrt {3 \times 7} }}{{\sqrt {2 \times 2} }} $
When the term is multiplied with the same integer it gives us the square of the term.
$ \sqrt {\dfrac{{21}}{4}} = \dfrac{{\sqrt {3 \times 7} }}{{\sqrt {{2^2}} }} $
Square and square-root cancel each other in the denominator of the above expression on the right hand side of the equation.
$ \sqrt {\dfrac{{21}}{4}} = \dfrac{{\sqrt {3 \times 7} }}{2} $
Also, the terms on the numerator of the above expression on the right are not the perfect square and so the simplified form can be re-written as –
$ \sqrt {\dfrac{{21}}{4}} = \dfrac{{\sqrt {21} }}{2} $
This is the required solution.
So, the correct answer is “$ \dfrac{{\sqrt {21} }}{2} $”.
Note: Be clear with the concepts of squares and square-roots. Square is defined as the number multiplied with itself. Square is the product of same number twice in such way as $ {n^2} = n \times n $ for Example square of $ 2 $ is $ {2^2} = 2 \times 2 $ simplified form of squared number is given by $ {2^2} = 2 \times 2 = 4 $ and square-root is denoted by $ \sqrt {{n^2}} = \sqrt {n \times n} $ For Example: $ \sqrt {{2^2}} = \sqrt 4 = 2 $
Complete step by step solution:
Take the given expression: $ \sqrt {\dfrac{{21}}{4}} $
Find the prime factors of the terms in the above expression. Factors of any term are terms which when multiplied together then it gives us the original term.
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 defined as the numbers which are greater than $ 1 $ and which are also not the product of any two smaller natural numbers.
$ \sqrt {\dfrac{{21}}{4}} = \sqrt {\dfrac{{3 \times 7}}{{2 \times 2}}} $
When the term in division is whole square-rooted we can rewrite by applying the square-root of the numerator upon the square of the denominator part.
$ \sqrt {\dfrac{{21}}{4}} = \dfrac{{\sqrt {3 \times 7} }}{{\sqrt {2 \times 2} }} $
When the term is multiplied with the same integer it gives us the square of the term.
$ \sqrt {\dfrac{{21}}{4}} = \dfrac{{\sqrt {3 \times 7} }}{{\sqrt {{2^2}} }} $
Square and square-root cancel each other in the denominator of the above expression on the right hand side of the equation.
$ \sqrt {\dfrac{{21}}{4}} = \dfrac{{\sqrt {3 \times 7} }}{2} $
Also, the terms on the numerator of the above expression on the right are not the perfect square and so the simplified form can be re-written as –
$ \sqrt {\dfrac{{21}}{4}} = \dfrac{{\sqrt {21} }}{2} $
This is the required solution.
So, the correct answer is “$ \dfrac{{\sqrt {21} }}{2} $”.
Note: Be clear with the concepts of squares and square-roots. Square is defined as the number multiplied with itself. Square is the product of same number twice in such way as $ {n^2} = n \times n $ for Example square of $ 2 $ is $ {2^2} = 2 \times 2 $ simplified form of squared number is given by $ {2^2} = 2 \times 2 = 4 $ and square-root is denoted by $ \sqrt {{n^2}} = \sqrt {n \times n} $ For Example: $ \sqrt {{2^2}} = \sqrt 4 = 2 $
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