Question

How do you simplify square root of 24 + square root of 54? $\sqrt{24}+\sqrt{54}$
 (a) Using algebraic properties
(b) Using linear formulas
(c) Using trigonometric formulas
(d) None of these

Answer 1

Hint: We are given a problem where we have to simplify a given term to get the solution. Here we will start with simplifying each of the numbers by using the prime factorisation and writing down the factors to obtain simpler form. Then we will add them and then simplify we are getting our desired result.

Complete step by step solution:
According to the problem, to start with we are left with finding the square root of 24 + square root of 54.
So, let us start with the first number which is 24.
Now, let us perform the prime factorization for 24 as below,
$\begin{align}
  & 2\left| \!{\underline {\,
  24 \,}} \right. \\
 & 2\left| \!{\underline {\,
  12 \,}} \right. \\
 & 2\left| \!{\underline {\,
  6 \,}} \right. \\
 & 3\left| \!{\underline {\,
  3 \,}} \right. \\
 & 1 \\
\end{align}$
Hence, 24 can be written as $3\times 2\times 2\times 2$ .
So, if we try to write $\sqrt{24}$ we get, $\sqrt{3\times 2\times 2\times 2}$.
As there are two 2’s inside the root term, one of them can be taken out.
So, 24 can be written as, $2\sqrt{6}$.
Again, start with the second number which is 54. Prime factorisation is as below,
$\begin{align}
  & 2\left| \!{\underline {\,
  54 \,}} \right. \\
 & 3\left| \!{\underline {\,
  27 \,}} \right. \\
 & 3\left| \!{\underline {\,
  9 \,}} \right. \\
 & 3\left| \!{\underline {\,
  3 \,}} \right. \\
 & 1 \\
\end{align}$
Hence, we can write 54 as $3\times 2\times 3\times 3$ .
So, if we try to write $\sqrt{54}$ we get, $\sqrt{3\times 2\times 3\times 3}$.
As there are two 3’s inside the root term, one of them can be taken out.
So, 54 can be written as, $3\sqrt{6}$.
Hence, $\sqrt{24}+\sqrt{54}$ can be written as, $2\sqrt{6}+3\sqrt{6}=5\sqrt{6}$. So, only using algebraic properties we have got our answer.

So, the correct answer is “Option a”.

Note: Here in this problem we are adding $3\sqrt{6}$ and $2\sqrt{6}$ in the last step. We have to keep that in mind that the terms are having the same numbers inside the root term and only that is why we are being able to add them altogether. If the terms inside the roots were different then it would not have been possible to add them.