Question

What is the square root of $\dfrac{35}{36}$?

Answer 1

Hint: Use the prime factorization method to find the square root of the denominator 36. Write it as the product of its prime factors and form pairs of identical factors such that the exponent becomes 2. Apply the formula ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$ to simplify. To find the square root of the numerator 35 use the long division method.

Complete step by step solution:
Here we have been asked to find the square root of $\dfrac{35}{36}$. Let us assume the square root of the given fraction is E, so we have,
$\Rightarrow E=\sqrt{\dfrac{35}{36}}$
Now writing the given numbers in the numerator and the denominator as the product of its primes we get,
$\Rightarrow E=\sqrt{\dfrac{5\times 7}{2\times 2\times 3\times 3}}$
Forming the pairs of identical factors and writing then in the exponent form with exponent equal to 2 we get,
$\Rightarrow E=\sqrt{\dfrac{5\times 7}{{{2}^{2}}\times {{3}^{2}}}}$
Using the identity ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$ we get,
\[\begin{align}
  & \Rightarrow E=\dfrac{\sqrt{5\times 7}}{{{\left( {{2}^{2}}\times {{3}^{2}} \right)}^{\dfrac{1}{2}}}} \\
 & \Rightarrow E=\dfrac{\sqrt{5\times 7}}{\left( {{2}^{2\times \dfrac{1}{2}}}\times {{3}^{2\times \dfrac{1}{2}}} \right)} \\
 & \Rightarrow E=\dfrac{\sqrt{5\times 7}}{2\times 3} \\
 & \Rightarrow E=\dfrac{\sqrt{5\times 7}}{6} \\
\end{align}\]
As we can see that we cannot form the pairs of identical prime factors in the numerator, so we need to use the long division method to find its square root. Let us find the root up to 2 places of decimal with the help of steps given below.
Step (1): Starting from the rightmost digit we will form pairs of two digits till all the digits are paired or a single digit is left at the leftmost place, here we have only two digits in 35 so they will get paired.
Step (2): We will select a number whose square will be less than or equal to 35. Here we have to select 5. Now, we will write 5 as the divisor and the quotient and subtract the product from 35. The remainder will be 10. Below the divisor’s place we will take the sum 5 + 5 = 10.
\[\begin{align}
  & \,\,\,\,\,\,\,\,\,\,\,\,5 \\
 & \begin{matrix}
   \,\,\,\,\,\,5 \\
   \,\,+5 \\
   \overline{\,\,\,\,10} \\
\end{matrix}\left| \!{\overline {\,
 \begin{align}
  & \,\,\,\,\,\,\,35 \\
 & \,\,\,-25 \\
 & \overline{\,\,\,\,\,\,\,10} \\
\end{align} \,}} \right. \\
\end{align}\]
Step (3): Now, there is no more pair of digits in the dividend to be placed beside the remainder 10, so we will take a decimal point after 5 in the quotient and put two 0’s after 10 in the remainder.
\[\begin{align}
  & \,\,\,\,\,\,\,\,\,\,\,\,5. \\
 & \begin{matrix}
   \,\,\,\,\,\,5 \\
   \,\,+5 \\
   \overline{\,\,\,\,10} \\
\end{matrix}\left| \!{\overline {\,
 \begin{align}
  & \,\,\,\,\,\,\,35 \\
 & \,\,\,-25 \\
 & \overline{\,\,\,\,\,\,\,1000} \\
\end{align} \,}} \right. \\
\end{align}\]
Step (4): Now, we need to select a digit after 10 in the divisor such that the product of this new number and the selected number will be less than 1000. Here we have to select 9. Now the new divisor will become 109 + 9 = 118, the remainder left will be 1000 – 981 = 19 and the new quotient will be 5.9.
\[\begin{align}
  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5.9 \\
 & \begin{matrix}
   \,\,\,\,\,\,\,\,5 \\
   \,\,\,\,+5 \\
   \overline{\,\,\,\,\,\,109} \\
   \,\,\,\,\,\,\,+9 \\
   \overline{\,\,\,\,\,\,118} \\
\end{matrix}\left| \!{\overline {\,
 \begin{align}
  & \,\,\,\,\,\,\,35 \\
 & \,\,\,-25 \\
 & \overline{\,\,\,\,\,\,\,1000} \\
 & \,\,\,\,\,\,-981 \\
 & \overline{\,\,\,\,\,\,\,\,\,\,\,\,\,19} \\
\end{align} \,}} \right. \\
\end{align}\]
Step (5): Again we will put two 0’s after the remainder so that the new dividend becomes 1900. Now, we need to select a digit after 118 in the divisor such that its product with the selected digit will be less than 1900. Here we need to select 1. Therefore, the new divisor will be 1181 + 1 = 1182, the remainder left will be 1900 – 1181 = 719 and the new quotient will be 5.91.
\[\begin{align}
  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5.91 \\
 & \begin{matrix}
   \,\,\,\,\,\,\,5 \\
   \,\,\,+5 \\
   \overline{\,\,\,\,\,\,\,\,\,\,\,\,109} \\
   \,\,\,\,\,\,\,\,\,\,\,\,+9 \\
   \overline{\,\,\,\,\,\,\,\,\,1181} \\
   \,\,\,\,\,\,\,\,\,\,\,\,+1 \\
   \overline{\,\,\,\,\,\,\,\,\,1182} \\
\end{matrix}\left| \!{\overline {\,
 \begin{align}
  & \,\,\,\,\,\,\,35 \\
 & \,\,\,-25 \\
 & \overline{\,\,\,\,\,\,\,1000} \\
 & \,\,\,\,\,-981 \\
 & \overline{\,\,\,\,\,\,\,\,\,\,\,\,1900} \\
 & \,\,\,\,\,\,\,\,-1181 \\
 & \overline{\,\,\,\,\,\,\,\,\,\,\,\,\,\,719} \\
\end{align} \,}} \right. \\
\end{align}\]
Therefore, the square root of 35 is 5.91 correct up to two places of decimal. So substituting this value in the expression ‘E’ we get,
\[\begin{align}
  & \Rightarrow E=\dfrac{5.91}{6} \\
 & \therefore E=0.985 \\
\end{align}\]

Hence, the square root of the given fraction is 0.985.

Note: You must remember the long division method to find the square root because this method is very useful in determining the square roots of numbers which are not perfect squares. In case of a number containing even number of digits all the digits will get paired while in case the number contains odd number of digits a digit will be left unpaired at the highest place value.