Question

How do you simplify $\dfrac{1}{\sqrt{32}}$ ?

Answer 1

Hint: To simplify the expression given above in which square root is given in the denominator. We are going to rationalize the denominator by multiplying and dividing the square root term written in the denominator. Then we can eliminate the square root in the denominator and hence the given expression will be simplified.

Complete step by step solution:
The expression given in the above problem is as follows:
$\dfrac{1}{\sqrt{32}}$
To simplify the above expression, we are going to eliminate the square root written in the denominator by rationalizing this expression by multiplying and dividing this fraction by $\sqrt{32}$ and we get,
$\begin{align}
  & \Rightarrow \dfrac{1}{\sqrt{32}}\times \dfrac{\sqrt{32}}{\sqrt{32}} \\
 & \Rightarrow \dfrac{\sqrt{32}}{32} \\
\end{align}$
Now, to simplify it further we are going to take the square root of 32. For that we are going to find the prime factorization of 32.
Prime factorization of 32 is as follows:
$32=2\times 2\times 2\times 2\times 2$
Now, to remove the square root we are going to write the above prime factorization as ${{2}^{4}}\times 2$. Now, putting this prime factorization in place of 32 written in the square root we get,
$\begin{align}
  & \Rightarrow \dfrac{\sqrt{{{2}^{4}}\times 2}}{32} \\
 & \Rightarrow \dfrac{{{2}^{2}}\sqrt{2}}{32} \\
\end{align}$
$\Rightarrow \dfrac{4\sqrt{2}}{32}$
In the above expression, 32 will be cancelled out by 4 to 8 times and we get,
$\Rightarrow \dfrac{\sqrt{2}}{8}$
Hence, we have simplified the above expression to $\dfrac{\sqrt{2}}{8}$.

Note: So, whenever you are asked to simplify the expressions which have square root terms in the denominator just like the one shown above then rationalize the denominator. Similarly, if you have denominator like given below:
$\begin{align}
  & \dfrac{1}{2+\sqrt{3}}; \\
 & \dfrac{1}{2-\sqrt{3}} \\
\end{align}$
Now, to simplify the above expressions we are going to rationalize the denominator as follows:
Rationalizing the above two expressions, we are going to multiply and divide the conjugate of the denominator. So, the conjugate of $2+\sqrt{3}$ is $2-\sqrt{3}$ and for $2-\sqrt{3}$ is $2+\sqrt{3}$. So, we are going to multiply and divide these conjugates respectively and we get,
\[\begin{align}
  & \dfrac{1}{2+\sqrt{3}}\times \dfrac{2-\sqrt{3}}{\left( 2-\sqrt{3} \right)}; \\
 & \dfrac{1}{2-\sqrt{3}}\times \dfrac{2+\sqrt{3}}{\left( 2+\sqrt{3} \right)} \\
\end{align}\]
Then further simplifying the above two expressions we get,
$\begin{align}
  & \dfrac{2-\sqrt{3}}{4-3}=2-\sqrt{3}; \\
 & \dfrac{2+\sqrt{3}}{4-3}=2+\sqrt{3} \\
\end{align}$