Question

Simplify the expression $\dfrac{\sqrt{24}}{8}+\dfrac{\sqrt{54}}{9}$ .

Answer 1

Hint: To simplify $\dfrac{\sqrt{24}}{8}+\dfrac{\sqrt{54}}{9}$ , we have to write the prime factors of 24 and 54. Then, we have to take the square root of the powers of 2. We then have to simplify the result. We have to take the LCM of the denominator and add the resultant expression.

Complete step by step answer:
We have to simplify $\dfrac{\sqrt{24}}{8}+\dfrac{\sqrt{54}}{9}$ . Let us write 24 and 54 in its prime factorization form.
$\begin{align}
  & 2\left| \!{\underline {\,
  24 \,}} \right. \\
 & 2\left| \!{\underline {\,
  12 \,}} \right. \\
 & 2\left| \!{\underline {\,
  6 \,}} \right. \\
 & 3\left| \!{\underline {\,
  3 \,}} \right. \\
 & \text{ }\text{ }\text{ }1 \\
\end{align}$
Therefore, we can write 24 as $2\times 2\times 2\times 3={{2}^{2}}\times 2\times 3$ .
Now, let us find the prime factors of 54.
$\begin{align}
  & 2\left| \!{\underline {\,
  54 \,}} \right. \\
 & 3\left| \!{\underline {\,
  27 \,}} \right. \\
 & 3\left| \!{\underline {\,
  9 \,}} \right. \\
 & 3\left| \!{\underline {\,
  3 \,}} \right. \\
 & \text{ }\text{ }\text{ }1 \\
\end{align}$
Therefore, we can write 54 as $2\times 3\times 3\times 3=2\times 3\times {{3}^{2}}$ .
Hence, we can write the given expression as
$\Rightarrow \dfrac{\sqrt{24}}{8}+\dfrac{\sqrt{54}}{9}=\dfrac{\sqrt{{{2}^{2}}\times 2\times 3}}{8}+\dfrac{\sqrt{2\times 3\times {{3}^{2}}}}{9}$
We know that $\sqrt[n]{ab}=\sqrt[n]{a}\times \sqrt[n]{b}$ . Therefore, we can write the above equation as
$\Rightarrow \dfrac{\sqrt{24}}{8}+\dfrac{\sqrt{54}}{9}=\dfrac{\sqrt{{{2}^{2}}}\times \sqrt{2\times 3}}{8}+\dfrac{\sqrt{{{3}^{2}}}\times \sqrt{2\times 3}}{9}$
We know that $\sqrt[n]{{{a}^{n}}}=a$ . Therefore, we can write the above equation as
$\begin{align}
  & \Rightarrow \dfrac{\sqrt{24}}{8}+\dfrac{\sqrt{54}}{9}=\dfrac{2\times \sqrt{2\times 3}}{8}+\dfrac{3\times \sqrt{2\times 3}}{9} \\
 & \Rightarrow \dfrac{\sqrt{24}}{8}+\dfrac{\sqrt{54}}{9}=\dfrac{2\sqrt{6}}{8}+\dfrac{3\sqrt{6}}{9} \\
\end{align}$
Let us cancel the common factor from the numerator and denominator of each term.
$\Rightarrow \dfrac{\sqrt{24}}{8}+\dfrac{\sqrt{54}}{9}=\dfrac{\require{cancel}\cancel{2}\sqrt{6}}{{{\require{cancel}\cancel{8}}^{4}}}+\dfrac{\require{cancel}\cancel{3}\sqrt{6}}{{{\require{cancel}\cancel{9}}^{3}}}$
We can write the result of the above simplification as
$\Rightarrow \dfrac{\sqrt{24}}{8}+\dfrac{\sqrt{54}}{9}=\dfrac{\sqrt{6}}{4}+\dfrac{\sqrt{6}}{3}$
Now, we have to take the LCM of the denominator. We know that LCM(4,3) is 12.
\[\begin{align}
  & \Rightarrow \dfrac{\sqrt{24}}{8}+\dfrac{\sqrt{54}}{9}=\dfrac{\sqrt{6}\times 3}{4\times 3}+\dfrac{\sqrt{6}\times 4}{3\times 4} \\
 & \Rightarrow \dfrac{\sqrt{24}}{8}+\dfrac{\sqrt{54}}{9}=\dfrac{3\sqrt{6}}{12}+\dfrac{4\sqrt{6}}{12} \\
 & \Rightarrow \dfrac{\sqrt{24}}{8}+\dfrac{\sqrt{54}}{9}=\dfrac{3\sqrt{6}+4\sqrt{6}}{12} \\
\end{align}\]
Let us add the roots.
\[\Rightarrow \dfrac{\sqrt{24}}{8}+\dfrac{\sqrt{54}}{9}=\dfrac{7\sqrt{6}}{12}\]
Therefore, the value of $\dfrac{\sqrt{24}}{8}+\dfrac{\sqrt{54}}{9}$ is \[\dfrac{7\sqrt{6}}{12}\] .

Note: Students must deeply understand to take the roots of the numbers and the methods associated with it. They must be thorough with the rules of the radicles. Students must know to solve fractional expressions. We can also solve the given expression in an alternate method.