Question

How do you find the square root of \[\dfrac{{81}}{{16}}\]?

Answer 1

Hint:In the above question, is based on the concept of taking square root. Given is the square root of a fraction where we need to calculate the square root of numerator and denominator both and the numbers given are already squares of a number therefore it is easy to calculate it.

Complete step by step solution:
Square root of a fraction can be explained as suppose the square root of a fraction \[\dfrac{x}{a}\] is that fraction \[\dfrac{y}{a}\]which when multiplied by itself gives the fraction of \[\dfrac{x}{a}\].

If x and a are numbers which already squares of a number then we can write it as,
$\sqrt {\dfrac{x}{a}} = \dfrac{{\sqrt x }}{{\sqrt a }}$

Here further the square root is separated and calculates the square root of numerator and denominator separately.

So, the given fraction is \[\dfrac{{81}}{{16}}\].Further by taking square root of the whole fraction we get,
\[\sqrt {\dfrac{{81}}{{16}}} \]

The next step will be to separate the square root sign on the fraction’s numerator and denominator both.

So, we can write it as,
\[\dfrac{{\sqrt {81} }}{{\sqrt {16} }}\]

We can see that 81 is the perfect square of the number 9 and the number in the denominator 16 is the perfect square of the number 4. So, substituting we get,
\[\dfrac{9}{4}\]

Therefore, this is the solution for a fraction of the number.

Note: An alternative method to solve this is by using prime factorisation . In this method the numbers in numerator and denominator should have its factors written then further taking square root.

For the number 81 we get \[\sqrt {81} = \sqrt {3 \times 3 \times 3 \times 3} \] and for 16 we get \[\sqrt {16} = \sqrt {4 \times 4 \times 4 \times 4} \]