Question

How do you simplify the square root of $ \dfrac{{45}}{{125}} $ ?

Answer 1

Hint: In order to write the expression into the simplest form, factorize the base part of the value inside the square root such that it contains perfect squares in it . The square root is related to figuring out what should be the number which when multiplied by itself is equal to the number under the square root symbol $ \sqrt {} $ . This symbol is known as radical. Since in our case we have the largest factor as 125 and 45 , so we will use here prime factorisation of 125 and 45 and pull out non- radical terms or perfect squares from the inside of the square root .

Complete step-by-step answer:
We have to find $ \sqrt {\dfrac{{45}}{{125}}} = \dfrac{{\sqrt {45} }}{{\sqrt {125} }} $ ---(1)
If we see the numbers under square root $ \dfrac{{45}}{{125}} $ both the numerator and denominator are composite numbers . We are going to make and choose the factor pairs by doing prime factorisation of the numbers but the factors should also be the perfect square .
The factorisation of 125 is as follows = $ 5 \times 5 \times 5 $ .and similarly the factorisation of 45 is $ 5 \times 3 \times 3 $
And the number 5 is not a perfect square but we can just only factorise it $ 5 \times 1 $ . $ $
The factorisation can be written as $ 125 = 5 \times 5 \times 5 $ .
This can also be rewritten as two raised to the power of 2 which makes the square which will pull out of the $ 125 = {5^2} \times 5 $ .
So , we will put the factorisation of 140 under square root and take the perfect square out for simplifying the given number .
 $ \sqrt {125} = \sqrt {{5^2} \times 5} $
As 2 raised to the power of five is regarded as square or the number 25 which itself is a perfect square by the definition of square root that states the number which when multiplied by itself is equal to the number under the square root is being implemented and satisfied .
We will pull out the number 5 raised to the power two when pulled out will makes the power unity that is 1 .
 $ \sqrt {125} = 5\sqrt 5 $
Similarly if we look at $ \sqrt {45} = \sqrt {{3^2} \times 5} = 3\sqrt 5 $
Putting the value of $ \sqrt {45} $ and $ \sqrt {125} $ in equation (1)
 $ \dfrac{{\sqrt {45} }}{{\sqrt {125} }} = \dfrac{{3\sqrt 5 }}{{5\sqrt 5 }} $
Cancelling $ \sqrt 5 $ as it is common in both numerator and denominator
 $ \dfrac{{\sqrt {45} }}{{\sqrt {125} }} = \dfrac{3}{5} $
Therefore the result of square root of $ \dfrac{{45}}{{125}} $ is equal to $ \dfrac{3}{5} $ .
So, the correct answer is “$ \dfrac{3}{5} $”.

Note: 1. Don’t Forgot to cross check the answer.
2.Factorise the number inside the square root properly to get knowledge of every perfect square.
3.Square root can be written as $ \sqrt {} = {\left( {} \right)^{\dfrac{1}{2}}} $ .