Question

Find the value of $\sqrt {24} \div \sqrt[3]{{200}}$ .

Answer 1

Hint: In the given problem, first we will find the prime factorization of numbers $24$ and $200$. Then, we will use the law of exponents which is given by ${\left( {{a^m}} \right)^n} = {a^{m \times n}}$. Also we will use law of exponents which is given by $\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}$ to find the required value.

Complete step-by-step answer:
In this problem, we have to find the value of $\sqrt {24} \div \sqrt[3]{{200}}$. We know that $\sqrt {24} $ can be written as ${\left( {24} \right)^{\dfrac{1}{2}}}$ and $\sqrt[3]{{200}}$ can be written as ${\left( {200} \right)^{\dfrac{1}{3}}}$. That is, we have to find the value of ${\left( {24} \right)^{\dfrac{1}{2}}} \div {\left( {200} \right)^{\dfrac{1}{3}}}$. Let us find the prime factorization of the number$24$.

$2$	$24$
$2$	$12$
$2$	$6$
$3$	$3$
	$1$

So, we can say that the prime factorization of the number $24$ is given by $24 = 2 \times 2 \times 2 \times 3$. That is, $24 = {2^3} \times 3$. Now we are going to find ${\left( {24} \right)^{\dfrac{1}{2}}}$ by using the law of exponents. So, we can write
${\left( {24} \right)^{\dfrac{1}{2}}} = {\left( {{2^3} \times 3} \right)^{\dfrac{1}{2}}}$
$ \Rightarrow {\left( {24} \right)^{\dfrac{1}{2}}} = {\left( {{2^3}} \right)^{\dfrac{1}{2}}} \times {\left( 3 \right)^{\dfrac{1}{2}}}\quad \left[ {\because {{\left( {ab} \right)}^m} = {a^m} \times {b^m}} \right]$
$ \Rightarrow {\left( {24} \right)^{\dfrac{1}{2}}} = {2^{\dfrac{3}{2}}} \times {3^{\dfrac{1}{2}}} \cdots \cdots \left( 1 \right)\quad \left[ {\because {{\left( {{a^m}} \right)}^n} = {a^{m \times n}}} \right]$
Now we are going to find the prime factorization of the number $200$.

$2$	$200$
$2$	$100$
$2$	$50$
$5$	$25$
$5$	$5$
	$1$

So, we can say that the prime factorization of the number $200$ is given by $200 = 2 \times 2 \times 2 \times 5 \times 5$. That is, $\left( {200} \right) = {2^3} \times {5^2}$. Now we are going to find ${\left( {200} \right)^{\dfrac{1}{3}}}$ by using the law of exponents. So, we can write
${\left( {200} \right)^{\dfrac{1}{3}}} = {\left( {{2^3} \times {5^2}} \right)^{\dfrac{1}{3}}}$
$ \Rightarrow {\left( {200} \right)^{\dfrac{1}{3}}} = {\left( {{2^3}} \right)^{\dfrac{1}{3}}} \times {\left( {{5^2}} \right)^{\dfrac{1}{3}}}\quad \left[ {\because {{\left( {ab} \right)}^m} = {a^m} \times {b^m}} \right]$
$ \Rightarrow {\left( {200} \right)^{\dfrac{1}{3}}} = 2 \times {5^{\dfrac{2}{3}}} \cdots \cdots \left( 2 \right)\quad \left[ {\because {{\left( {{a^m}} \right)}^n} = {a^{m \times n}}} \right]$
Now we are going to find the value of ${\left( {24} \right)^{\dfrac{1}{2}}} \div {\left( {200} \right)^{\dfrac{1}{3}}}$ from $\left( 1 \right)$ and $\left( 2 \right)$. So, we can write
${\left( {24} \right)^{\dfrac{1}{2}}} \div {\left( {200} \right)^{\dfrac{1}{3}}}$
$ = \dfrac{{{2^{\dfrac{3}{2}}} \times {3^{\dfrac{1}{2}}}}}{{2 \times {5^{\dfrac{2}{3}}}}}$

$ = \dfrac{{{2^{^{\dfrac{3}{2} - 1}}} \times {3^{\dfrac{1}{2}}}}}{{{5^{\dfrac{2}{3}}}}}\quad \quad \left[ {\because \dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}} \right]$

$ = \dfrac{{{2^{\dfrac{1}{2}}} \times {3^{\dfrac{1}{2}}}}}{{{5^{\dfrac{2}{3}}}}}$
$ = \dfrac{{\sqrt 2 \times \sqrt 3 }}{{{{\left( {{5^2}} \right)}^{\dfrac{1}{3}}}}}$
$ = \dfrac{{\sqrt 6 }}{{\sqrt[3]{{25}}}}$
Hence, the required value is $\dfrac{{\sqrt 6 }}{{\sqrt[3]{{25}}}}$ or $\dfrac{{{6^{\dfrac{1}{2}}}}}{{{{25}^{\dfrac{1}{3}}}}}$.

Note: In this problem, we can find the required value by direct simplification also. $\sqrt {24} $ can be written as $\sqrt {4 \times 6} = 2\sqrt 6 $ and $\sqrt[3]{{200}}$ can be written as $\sqrt[3]{{8 \times 25}} = 2\sqrt[3]{{25}}$. So, we can write $\dfrac{{\sqrt {24} }}{{\sqrt[3]{{200}}}} = \dfrac{{2\sqrt 6 }}{{2\sqrt[3]{{25}}}} = \dfrac{{\sqrt 6 }}{{\sqrt[3]{{25}}}}$. Here we can see that answer remains the same.