Question

How do you simplify $\sqrt{121}$ ?

Answer 1

Hint: We are given $\sqrt{121}$ , we have to simplify. We will learn how to simplify such types of expressions. With some examples to get the understanding then we work on our $\sqrt{121}$ , we start with prime factorization of 121 and then we take a pair out of the root having all un paired inside it and then we simplify and solve.

Complete step by step answer:
We are given a equation a term with square root, inside it we have 121 to learn how to simplicity such question we will learn some example related to this, to solve such problem where we have radical $\left( \sqrt{{}} \right)$ in the numerator only and denominator is simply one, this type of question are mainly based on prime factorization to simplify.
Like we consider we have $\sqrt{8}$
So we have 8 inside the square root, we factor 8 into its prime factor so we get $8=2\times 2\times 2$ .
So, we get –
$\sqrt{8}=\sqrt{2\times 2\times 2}$
Now properties of square root is only a pair at term can be taken out of it as we can have 3 pieces of 2, so 2 pieces of 2 will make one pair so it will come out and the remaining one will stay inside the root.
So,
$\begin{align}
  & \sqrt{8}=\sqrt{2\times 2\times 2} \\
 & =2\times \sqrt{2} \\
\end{align}$
So we get –
$\sqrt{8}=2\sqrt{2}$
Similarly say if we have $\sqrt{16}$
So,
$\sqrt{16}=\sqrt{2\times 2\times 2\times 2}$
Here we are getting fully paired up so all terms will come out and there will be nothing inside root.
$\begin{align}
  & \sqrt{16}=2\times 2 \\
 & =4 \\
\end{align}$
So simplification of $\sqrt{16}$ is 4
Now, we were on our problem, we have $\sqrt{121}$
So we start our solution by finding the prime factor of 121
So we get that –
$121=11\times 11$
We can see that in the prime factor of 121, we have 11 making a pair so 11 will be taken out, apart from 11 there is nothing else inside.
So,
$\sqrt{121}=\sqrt{11\times 11}$
So, $\sqrt{121}=11$
Hence, we get –
$\sqrt{121}$ is 11.

Note:
Always remember only prime factorization is done to get the square root of any term or like if we have problem of cube root $\left( \sqrt[3]{{}} \right)$ then we will make the pair of 3 inside the radical term who make pair of 3 will taken out of cube root similarly for nth root $\left( \sqrt[n]{{}} \right)$ we make pair of n term at once and those who do make pair will be taken out.