Question

Write the simplest rationalizing factor of $\left( i \right)\sqrt {32} \left( {ii} \right)\sqrt {72} \left( {iii} \right)\sqrt[3]{5}\left( {iv} \right)\sqrt[5]{9}\left( v \right)\sqrt[3]{{135}}$

Answer 1

Hint: We are given few surds and we need to rationalize it using a factor that is known as the rationalizing factor. In problems like these, we need to analyze the type of root and the number inside the root. We need to check whether the number can be written as any multiple and simplify it accordingly. Finally, after simplification, we need to find the simplest number which will make it a rational number.

Complete step-by-step answer:
We are given a few surds and we need to find the smallest rationalizing factor of the surds.
Let’s start with the first problem

$\left( i \right)\sqrt {32} $
Here first we need to analyze the number inside the root.
From the question, it is clearly understood the root is a square root.
The number inside the root is $32$
We know that $32$ can be written as $16 \times 2$
Hence, we get,
$\sqrt {32} = \sqrt {16 \times 2} $
We know that $\sqrt {16} = 4$
Using this we get,
$\sqrt {32} = 4\sqrt 2 $
Now here the number inside the root is $2$ . So now the only way to rationalize it is to multiply it by $\sqrt 2 $
Hence multiplying by $\sqrt 2 $, we get
$4\sqrt 2 \times \sqrt 2 = 4 \times 2 = 8$
Now the resulting number $8$ is a rational number.
Hence the rationalizing factor $\sqrt {32} $ is $\sqrt 2 $

$\left( {ii} \right)\sqrt {72} $
Here first we need to analyze the number inside the root.
From the question, it is clearly understood the root is the square root.
The number inside the root is $72$
We know that $72$ can be written as $36 \times 2$
Hence, we get,
$\sqrt {72} = \sqrt {36 \times 2} $
We know that $\sqrt {36} = 6$
Using this we get,
$\sqrt {72} = 6\sqrt 2 $
Now here the number inside the root is $2$ . So now the only way to rationalize it is to multiply it by $\sqrt 2 $
Hence multiplying by $\sqrt 2 $, we get
$6\sqrt 2 \times \sqrt 2 = 6 \times 2 = 12$
Now the resulting number $12$ is a rational number.
Hence the rationalizing factor $\sqrt {72} $ is $\sqrt 2 $

$\left( {iii} \right)\sqrt[3]{5}$
Here first we need to analyze the number inside the root.
From the question, it is clearly understood that the root is the cube root.
The number inside the root is $5$
We know that $5$ is a prime number so it cannot be written as any multiple
So now the only way to rationalize it is to multiply it by $\sqrt[3]{{25}}$ as this will make the number inside the root a cube number
Hence multiplying by $\sqrt[3]{{25}}$, we get
$\sqrt[3]{5} \times \sqrt[3]{{25}} = \sqrt[3]{{125}}$
We know that $125$ can be written as $5 \times 5 \times 5$
Hence, we get,
$\sqrt[3]{{125}} = \sqrt[3]{{5 \times 5 \times 5}} = 5$
Now the resulting number $5$ is a rational number.
Hence the rationalizing factor $\sqrt[3]{5}$ is $\sqrt[3]{{25}}$

$\left( {iv} \right)\sqrt[5]{9}$
Here first we need to analyze the number inside the root.
From the question, it is clearly understood the root is the fifth root.
The number inside the root is $9$
We know that $9$ can be written as $3 \times 3$
So now the only way to rationalize it is to multiply it by $\sqrt[5]{{27}}$
Hence multiplying by $\sqrt[5]{{27}}$, we get
$\sqrt[5]{{3 \times 3}} \times \sqrt[5]{{27}} = \sqrt[3]{{3 \times 3 \times 27}}$
We know that $27$ can be written as \[3 \times 3 \times 3\]
Hence, we get,
$\sqrt[5]{{3 \times 3 \times 27}} = \sqrt[5]{{3 \times 3 \times 3 \times 3 \times 3}} = 3$
Now the resulting number $3$ is a rational number.
Hence the rationalizing factor $\sqrt[5]{9}$ is $\sqrt[5]{{27}}$

$\left( v \right)\sqrt[3]{{135}}$
Here first we need to analyze the number inside the root.
From the question, it is clearly understood that the root is the cube root.
The number inside the root is $135$
We know that $135$ can be written as \[27 \times 5\]
Hence, we get,
\[\sqrt[3]{{135}} = \sqrt[3]{{27 \times 5}}\]
We know that $\sqrt[3]{{27}} = 3$
Using this we get,
\[\sqrt[3]{{135}} = 3\sqrt[3]{5}\]
So now the only way to rationalize it is to multiply it by $\sqrt[3]{{25}}$ as this will make the number inside the root a cube number
Hence multiplying by $\sqrt[3]{{25}}$, we get
$3\sqrt[3]{5} \times \sqrt[3]{{25}} = 3\sqrt[3]{{125}}$
We know that $125$ can be written as $5 \times 5 \times 5$
Hence, we get,
$3\sqrt[3]{{125}} = 3\sqrt[3]{{5 \times 5 \times 5}} = 3 \times 5 = 15$
Now the resulting number $15$ is a rational number.
Hence the rationalizing factor $\sqrt[3]{{135}}$ is $\sqrt[3]{{25}}$

Note: The student has to be careful while checking what kind of root is given and whether the number inside the root can be written accordingly. When it is square root, the number can be easily rationalized by multiplying it by itself once. But this cannot be followed if the root is cube root or anything else.