Question

How do you simplify $ {{36}^{\dfrac{1}{2}}} $ ?

Answer 1

Hint: We try to form the indices formula for the value 2. This is a square root of 36. We find the prime factorization of 36. Then we take one digit out of the two same number of primes. 36 is square of 6. Therefore, the square root gives the result of 6.

Complete step by step answer:
We need to find the value of the algebraic form of $ {{36}^{\dfrac{1}{2}}} $ . This is a square root form.
The given value is the form of indices. We are trying to find the root value of 36.
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 factorization 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 numbers of primes.
This means in the root value of $ 36=2\times 2\times 3\times 3 $, we will take out one 2 and one 3.
So, $ \sqrt{36}=\sqrt{2\times 2\times 3\times 3}=2\times 3=6 $ . Basically 36 is the multiplication of two sixes.
We can also use the theorem of indices \[{{\left( {{a}^{x}} \right)}^{y}}={{a}^{xy}}\]. We know that $ 36={{6}^{2}} $ .
We need to find $ {{36}^{\dfrac{1}{2}}} $ which gives $ {{36}^{\dfrac{1}{2}}}={{\left( {{6}^{2}} \right)}^{\dfrac{1}{2}}}={{6}^{2\times \dfrac{1}{2}}}=6 $ .
Therefore, the value of $ {{36}^{\dfrac{1}{2}}} $ is 6.

Note:
We can also use the variable form where we can take $ x={{36}^{\dfrac{1}{2}}} $ . But we need to remember that we can’t use the square on both sides of the equation $ x={{36}^{\dfrac{1}{2}}} $ as in that case we are taking an extra value of negative as in $ -6 $ as a root value. Then this linear equation becomes a quadratic equation.