Question

How do you simplify \[\sqrt {\dfrac{{72}}{9}} \] ?

Answer 1

Hint: In order to solve this question, we will simplify the given algebraic expression by using the formula \[\sqrt {\dfrac{a}{b}} = \dfrac{{\sqrt a }}{{\sqrt b }}\] .Then to simplify the square root of \[72\] and \[9\] we will do the prime factorization. After that we will pair the same prime factors. The unpaired ones will be multiplied inside the square root while the paired ones will be taken outside as a single number and multiplied outside. Hence, we will simplify and get the required answer.

Complete step by step answer:
The given algebraic expression is \[\sqrt {\dfrac{{72}}{9}} \]
It expresses the square root of the fraction that consists of number seventy- two as numerator and nine as denominator.
Now we know that
\[\sqrt {\dfrac{a}{b}} = \dfrac{{\sqrt a }}{{\sqrt b }}\]
Therefore, we can write the given expression as
\[\sqrt {\dfrac{{72}}{9}} = \dfrac{{\sqrt {72} }}{{\sqrt 9 }}{\text{ }} - - - \left( i \right)\]
Now let’s deal with the two roots separately.
Now let’s prime factorize the numbers that are in the square root
So, prime factors are as follows:
\[72 = 2 \times 2 \times 2 \times 3 \times 3\]
\[9 = 3 \times 3{\text{ }}\]
Now we have to pair the same prime factors.
\[ \Rightarrow 72 = \left( {2 \times 2} \right) \times 2 \times \left( {3 \times 3} \right){\text{ }}\]
\[ \Rightarrow 9 = \left( {3 \times 3{\text{ }}} \right)\]
Now on taking square root we get
\[ \Rightarrow \sqrt {72} = \sqrt {\left( {2 \times 2} \right) \times 2 \times \left( {3 \times 3} \right){\text{ }}} \]
\[ \Rightarrow \sqrt 9 = \sqrt {\left( {3 \times 3{\text{ }}} \right)} \]
So, on putting in the equation \[\left( i \right)\] , we get
\[\dfrac{{\sqrt {72} }}{{\sqrt 9 }} = \dfrac{{\sqrt {\left( {2 \times 2} \right) \times 2 \times \left( {3 \times 3} \right)} }}{{\sqrt {\left( {3 \times 3} \right)} }}\]
Now the unpaired ones will be multiplied inside the square root while the paired ones will be taken out as a single number from the square root.
Therefore, we get
\[\dfrac{{\sqrt {72} }}{{\sqrt 9 }} = \dfrac{{2 \times 3 \times \sqrt 2 }}{3}\]

On simplifying, we get
\[\dfrac{{\sqrt {72} }}{{\sqrt 9 }} = 2\sqrt 2 \]
Using equation \[\left( i \right)\] we get
\[\dfrac{{\sqrt {72} }}{{\sqrt 9 }} = 2\sqrt 2 \]
Hence, the simplified form of \[\sqrt {\dfrac{{72}}{9}} \] is \[2\sqrt 2 \].

Note:
For solving this type of question, we must know how to prime factorize the numbers. We can see that \[72\] is not a perfect square while \[9\] is a perfect square. We can solve this problem in an alternative method as
Here, \[9\] is a perfect square, that is \[\sqrt 9 = 3{\text{ }} - - - \left( i \right)\]
So, let’s write \[72\] as a product of two numbers where each or one of these numbers are perfect squares.
Therefore, we can write \[72\] as \[2 \times 36\]
\[ \Rightarrow \sqrt {72} = \sqrt {2 \times 36} \]
Now we know that
\[\sqrt {a \times b} = \sqrt a \times \sqrt b \]
Therefore, we get
\[ \Rightarrow \sqrt {72} = \sqrt 2 \times \sqrt {36} \]
We know that \[36\] is a perfect square, that is, \[36 = {6^2}\]
On taking square root, we will get
\[ \Rightarrow \sqrt {72} = \sqrt 2 \times 6{\text{ }} - - - \left( {ii} \right)\]
Therefore, using \[\left( i \right)\] and \[\left( {ii} \right)\] we get the final result as
\[\sqrt {\dfrac{{72}}{9}} = \dfrac{{\sqrt {72} }}{{\sqrt 9 }} = \dfrac{{6\sqrt 2 }}{3}\]
\[\sqrt {\dfrac{{72}}{9}} = 2\sqrt 2 \] which is the required result.