Is 36 a perfect square?

Hint: We try to form the indices formula for the value 2. This is finding square root of 36. We find the prime factorisation of 36. Then we take one digit out of the two same number of primes. There will be no odd number of primes remaining in the root which will prove that 36 is a perfect square.

Complete step-by-step solution:
We try to find the value of the algebraic form of $\sqrt{36}$. This is a square root form.
We know the theorem of indices \[{{a}^{\dfrac{1}{n}}}=\sqrt[n]{a}\]. Putting value 2 we get \[{{a}^{\dfrac{1}{2}}}=\sqrt{a}\].
We need to find the prime factorisation of the given number 36.
$\begin{align}
  & 2\left| \!{\underline {\,
  36 \,}} \right. \\
 & 2\left| \!{\underline {\,
  18 \,}} \right. \\
 & 3\left| \!{\underline {\,
  9 \,}} \right. \\
 & 3\left| \!{\underline {\,
  3 \,}} \right. \\
 & 1\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}$
Therefore, \[36=2\times 2\times 3\times 3\].
For finding the square root, we need to take one digit out of the two same number of primes.
This means in the cube root value of 36, we will take out one 2 and one 3 from the multiplication and there will remain no extra prime numbers.
We get \[\sqrt{36}=\sqrt{2\times 2\times 3\times 3}=2\times 3=6\].
Therefore, 36 is a perfect square and its square root is 6.

Note: We can also use the variable form where we can take $x=\sqrt{36}$. But we need to remember that we can’t use the square on both sides of the equation $x=\sqrt{36}$ as in that case we are taking one extra value as a root value. Then this linear equation becomes a quadratic equation.