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How do you find the square root of \[\dfrac{9}{144}\]?
Answer
452.7k+ views
Hint: Assume the given expression as ‘E’. Simplify this expression by cancelling the common factors. To do this, use the prime factorization method to write the given numerator and denominator as the product of their primes. Now, cancel the factors which are common in them. Take the square root of the simplified expression and form a group of two for the remaining factors. Cancel the exponent 2 and square root to get the answer.
Complete step-by-step answer:
Here, we have been provided with the expression \[\dfrac{9}{144}\] and we are asked to find its square root. Let us assume the given expression as E. So, we have,
\[\Rightarrow E=\dfrac{9}{144}\]
Now, we have to check if there are common factors for the numerator and denominator or not. If there are then we have to cancel them and simplify the expression ‘E’. To do this, we will write the given numbers as the product of their primes. So, we have,
\[\begin{align}
& \Rightarrow 9=3\times 3 \\
& \Rightarrow 144=2\times 2\times 2\times 2\times 3\times 3 \\
\end{align}\]
So, substituting these in expression E, we get,
\[\Rightarrow E=\dfrac{3\times 3}{2\times 2\times 2\times 2\times 3\times 3}\]
\[\Rightarrow E=\dfrac{1}{2\times 2\times 2\times 2}\] (on cancelling the common factors)
Now, taking square root both the sides, we get,
\[\Rightarrow \sqrt{E}=\sqrt{\dfrac{1}{2\times 2\times 2\times 2}}\]
As we can see that there are four 2’s, so on grouping two 2’s together, we get,
\[\Rightarrow \sqrt{E}=\sqrt{\dfrac{1}{\left( 2\times 2 \right)\times \left( 2\times 2 \right)}}\]
In exponent form it can be written as: -
\[\begin{align}
& \Rightarrow \sqrt{E}=\sqrt{\dfrac{1}{{{2}^{2}}\times {{2}^{2}}}} \\
& \Rightarrow \sqrt{E}={{\left( \dfrac{1}{{{2}^{2}}\times {{2}^{2}}} \right)}^{\dfrac{1}{2}}} \\
& \Rightarrow \sqrt{E}=\dfrac{1}{{{\left( {{2}^{2}}\times {{2}^{2}} \right)}^{\dfrac{1}{2}}}} \\
\end{align}\]
Using the formula \[{{a}^{m}}\times {{b}^{m}}={{\left( a\times b \right)}^{m}}\], we get,
\[\begin{align}
& \Rightarrow \sqrt{E}=\dfrac{1}{{{\left[ {{\left( 2\times 2 \right)}^{2}} \right]}^{\dfrac{1}{2}}}} \\
& \Rightarrow \sqrt{E}=\dfrac{1}{{{\left[ {{\left( 4 \right)}^{2}} \right]}^{\dfrac{1}{2}}}} \\
\end{align}\]
Using the formula: - \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\], we get,
\[\begin{align}
& \Rightarrow \sqrt{E}=\dfrac{1}{{{4}^{2\times \dfrac{1}{2}}}} \\
& \Rightarrow \sqrt{E}=\dfrac{1}{4} \\
\end{align}\]
Hence, the square root of \[\dfrac{9}{144}\] is \[\dfrac{1}{4}\].
Note: One may note that here we have simplified the given expression ‘E’ and then found the square root. You can also do the reverse process, that is, first find the square root and then simplify. You must know how to find prime factors of a number because in many cases the given number will be very large and in such cases the given number will be very large and in such a case we would have no option other than finding the prime factors. Note that here we were asked to find the square root and that is why we have formed a group of two identical factors. If we were to find a cube root then we would have formed a group of three identical factors.
Complete step-by-step answer:
Here, we have been provided with the expression \[\dfrac{9}{144}\] and we are asked to find its square root. Let us assume the given expression as E. So, we have,
\[\Rightarrow E=\dfrac{9}{144}\]
Now, we have to check if there are common factors for the numerator and denominator or not. If there are then we have to cancel them and simplify the expression ‘E’. To do this, we will write the given numbers as the product of their primes. So, we have,
\[\begin{align}
& \Rightarrow 9=3\times 3 \\
& \Rightarrow 144=2\times 2\times 2\times 2\times 3\times 3 \\
\end{align}\]
So, substituting these in expression E, we get,
\[\Rightarrow E=\dfrac{3\times 3}{2\times 2\times 2\times 2\times 3\times 3}\]
\[\Rightarrow E=\dfrac{1}{2\times 2\times 2\times 2}\] (on cancelling the common factors)
Now, taking square root both the sides, we get,
\[\Rightarrow \sqrt{E}=\sqrt{\dfrac{1}{2\times 2\times 2\times 2}}\]
As we can see that there are four 2’s, so on grouping two 2’s together, we get,
\[\Rightarrow \sqrt{E}=\sqrt{\dfrac{1}{\left( 2\times 2 \right)\times \left( 2\times 2 \right)}}\]
In exponent form it can be written as: -
\[\begin{align}
& \Rightarrow \sqrt{E}=\sqrt{\dfrac{1}{{{2}^{2}}\times {{2}^{2}}}} \\
& \Rightarrow \sqrt{E}={{\left( \dfrac{1}{{{2}^{2}}\times {{2}^{2}}} \right)}^{\dfrac{1}{2}}} \\
& \Rightarrow \sqrt{E}=\dfrac{1}{{{\left( {{2}^{2}}\times {{2}^{2}} \right)}^{\dfrac{1}{2}}}} \\
\end{align}\]
Using the formula \[{{a}^{m}}\times {{b}^{m}}={{\left( a\times b \right)}^{m}}\], we get,
\[\begin{align}
& \Rightarrow \sqrt{E}=\dfrac{1}{{{\left[ {{\left( 2\times 2 \right)}^{2}} \right]}^{\dfrac{1}{2}}}} \\
& \Rightarrow \sqrt{E}=\dfrac{1}{{{\left[ {{\left( 4 \right)}^{2}} \right]}^{\dfrac{1}{2}}}} \\
\end{align}\]
Using the formula: - \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\], we get,
\[\begin{align}
& \Rightarrow \sqrt{E}=\dfrac{1}{{{4}^{2\times \dfrac{1}{2}}}} \\
& \Rightarrow \sqrt{E}=\dfrac{1}{4} \\
\end{align}\]
Hence, the square root of \[\dfrac{9}{144}\] is \[\dfrac{1}{4}\].
Note: One may note that here we have simplified the given expression ‘E’ and then found the square root. You can also do the reverse process, that is, first find the square root and then simplify. You must know how to find prime factors of a number because in many cases the given number will be very large and in such cases the given number will be very large and in such a case we would have no option other than finding the prime factors. Note that here we were asked to find the square root and that is why we have formed a group of two identical factors. If we were to find a cube root then we would have formed a group of three identical factors.
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