Question

What is the simplest radical form of 53?

Answer 1

Hint: To find the simplest radical form of this number 53. We are going to take the square root of 53. And then we will try to write 53 in such a manner so that we can write 53 as some highest square of any number multiplied by some other number.

Complete step-by-step solution:
In the above problem, we have given a number 53 and we are asked to write the simplest radical form.
For that, we are going to write 53 as square root of 53 then writing the square root of 53 we get,
$\sqrt{53}$
Now, we are going to factorize the number 53. As 53 is a prime number so there are only two factors possible for this number 53, one of the factors is 1 and the other factor is 53. So, writing the factors for 53 we get,
$53=1\times 53$
Now, substituting the above factor form in place of 53 in $\sqrt{53}$ we get,
$\sqrt{1\times 53}$
In the above square root expression, we cannot take any of the two numbers 1 and 53 outside the square root because both of the two numbers don’t have any power which is even.
Hence, the simplest radical form of 53 is $\sqrt{53}$.

Note: Suppose instead of 53, you are asked to find the simplest radical form for 225. Then we are going to find the square root of 225. And the square root of 225 is equal to:
$\sqrt{225}$
Now, we are going to do the factorization of 225 and we get,
$225=1\times 3\times 5\times 3\times 5$
Now, we can write the two times same number multiplication as ${{3}^{2}}\And {{5}^{2}}$ because when the base is same then power gets added in the multiplication of two numbers.
Then the above factorization will look as follows:
$225={{3}^{2}}\times {{5}^{2}}$
Also, we can write the above equation as:
$\begin{align}
  & 225={{\left( 3\times 5 \right)}^{2}} \\
 & \Rightarrow 225={{15}^{2}} \\
\end{align}$
Substituting the above value of 225 in the square root of 225 we get,
$\begin{align}
  & \sqrt{{{15}^{2}}} \\
 & ={{15}^{2\times \dfrac{1}{2}}} \\
\end{align}$
In the exponent of the above number, 2 will be cancelled out from the numerator and the denominator and we get,
$\begin{align}
  & ={{15}^{1}} \\
 & =15 \\
\end{align}$