Question

How do you simplify $\sqrt{540}$?

Answer 1

Hint: We have been asked to simplify $\sqrt{540}$. So, we will first learn how to simplify such types of expressions, with the help of some examples. Then we will work on our problem, we will start with prime factorization of 84 and then we will take a pair out of the root leaving all unpaired ones inside it and then we will simplify and solve.

Complete answer:
We are given a term with square root and we have the number 540 inside it. To learn how to simplify such questions, we will take some examples and solve such problems where we have radicals in the numerator only and denominators simply one. These types of questions are based on prime factorization.
Let us consider $\sqrt{8}$, so here, we will factorize 8 into its prime factors, so we will get,
$8=2\times 2\times 2$
So, we can write,
  $\sqrt{8}=\sqrt{2\times 2\times 2}$
Now, the property of square root is that only a pair of terms can be taken out of it, as we can see that here we have three 2s, so we can take out only two 2s as a pair, and the remaining one 2 will be inside the square root itself. So, we get,
$\sqrt{8}=2\sqrt{2}$.
Similarly, say we have $\sqrt{16}$. So, we can express it as follows.
$\sqrt{16}=\sqrt{2\times 2\times 2\times 2}$
Here, we have two pairs of 2, hence, we can take them outside. So, we will get,
$\sqrt{16}=2\times 2=4$
Now, let us consider the term we have been asked to solve, that is, $\sqrt{540}$.
So, let us start by finding the prime factors of 540.
We get that the prime factors of 540 are 2, 3 and 5. So, we can write,
$540=2\times 2\times 3\times 3\times 3\times 5$
We can see that the factor of 540 has a pair of 2 and 3 and one 3 and 5 left unpaired. So, taking the pairs of 2 and 3 outside with 3 and 5 remaining inside the square root, we get,
$\begin{align}
  & \sqrt{540}=\sqrt{2\times 2\times 3\times 3\times 3\times 5} \\
 & =2\times 2\sqrt{3\times 5} \\
\end{align}$
On simplifying we get,
$\sqrt{540}=6\sqrt{15}$
Therefore, $\sqrt{540}=6\sqrt{15}$.

Note: We take only one factor out of the pair of same factors outside the square root. It means that if we have a pair of 2, then it becomes one 2 when taken outside the square root.
So, we need to be careful regarding this. Also, we need to perform prime factorization and consider the prime factors only. Suppose we have a factor like 4, 6 etc., we are not supposed to consider them, instead we should represent them as products of prime factors like 6 can be written products of 2 and 3.