How do you simplify $\dfrac{1}{{\sqrt {32} }}$?
Answer
Verified
450.9k+ views
Hint: This problem deals with simplifying the given radical expression. This is done by simplifying the radical expressions using the product and quotient rule for radicals. Use formulas involving radicals. Evaluate given square root and cube root functions. An algebraic expression that contains radicals is called a radical expression. We use the product and quotient rules to simplify them.
Complete step-by-step answer:
Given the radical expression $\dfrac{1}{{\sqrt {32} }}$, consider this expression as given below:
$ \Rightarrow \dfrac{1}{{\sqrt {32} }}$
Now rationalize the denominator by multiplying the given fraction with $\sqrt {32} $. That is multiplying and dividing the given fraction with $\sqrt {32} $, as shown below:
$ \Rightarrow \dfrac{1}{{\sqrt {32} }} \times \dfrac{{\sqrt {32} }}{{\sqrt {32} }}$
$ \Rightarrow \dfrac{{\sqrt {32} }}{{32}}$
Now simplifying the numerator which is $\sqrt {32} $ , as given below:
The $\sqrt {32} $ can be written as: $\sqrt {32} = \sqrt {8 \times 4} $
$ \Rightarrow \sqrt {32} = \sqrt {2 \times 2 \times 2 \times 2 \times 2} $
We know that 8, is the cube of 2, and whereas 4, is the square of 2.
$\therefore \sqrt {32} = 4\sqrt 2 $
Now substituting the value of $\sqrt {32} = 4\sqrt 2 $ in the numerator, as shown below:
$ \Rightarrow \dfrac{{\sqrt {32} }}{{32}} = \dfrac{{4\sqrt 2 }}{{32}}$
Now simplifying the numbers, in the numerator and denominator, which are outside the root, as shown below:
$ \Rightarrow \dfrac{{4\sqrt 2 }}{{32}} = \dfrac{{\sqrt 2 }}{8}$
So the value of the radical expression $\dfrac{1}{{\sqrt {32} }}$, is given by:
$\therefore \dfrac{1}{{\sqrt {32} }} = \dfrac{1}{8}\sqrt 2 $
Final Answer: The simplification of the radical expression of $\dfrac{1}{{\sqrt {32} }}$ is equal to $\dfrac{{\sqrt 2 }}{8}$.
Note:
Please note that if you want to multiply, first coefficients are multiplied with each other and the sub-radical amounts each other, placing the latter product under the radical sign common and the result is simplified.
But if you want to divide, then the coefficients are divided among themselves and sub-radical amounts each other, placing the latter quotient under the radical common and the result is simplified.
Complete step-by-step answer:
Given the radical expression $\dfrac{1}{{\sqrt {32} }}$, consider this expression as given below:
$ \Rightarrow \dfrac{1}{{\sqrt {32} }}$
Now rationalize the denominator by multiplying the given fraction with $\sqrt {32} $. That is multiplying and dividing the given fraction with $\sqrt {32} $, as shown below:
$ \Rightarrow \dfrac{1}{{\sqrt {32} }} \times \dfrac{{\sqrt {32} }}{{\sqrt {32} }}$
$ \Rightarrow \dfrac{{\sqrt {32} }}{{32}}$
Now simplifying the numerator which is $\sqrt {32} $ , as given below:
The $\sqrt {32} $ can be written as: $\sqrt {32} = \sqrt {8 \times 4} $
$ \Rightarrow \sqrt {32} = \sqrt {2 \times 2 \times 2 \times 2 \times 2} $
We know that 8, is the cube of 2, and whereas 4, is the square of 2.
$\therefore \sqrt {32} = 4\sqrt 2 $
Now substituting the value of $\sqrt {32} = 4\sqrt 2 $ in the numerator, as shown below:
$ \Rightarrow \dfrac{{\sqrt {32} }}{{32}} = \dfrac{{4\sqrt 2 }}{{32}}$
Now simplifying the numbers, in the numerator and denominator, which are outside the root, as shown below:
$ \Rightarrow \dfrac{{4\sqrt 2 }}{{32}} = \dfrac{{\sqrt 2 }}{8}$
So the value of the radical expression $\dfrac{1}{{\sqrt {32} }}$, is given by:
$\therefore \dfrac{1}{{\sqrt {32} }} = \dfrac{1}{8}\sqrt 2 $
Final Answer: The simplification of the radical expression of $\dfrac{1}{{\sqrt {32} }}$ is equal to $\dfrac{{\sqrt 2 }}{8}$.
Note:
Please note that if you want to multiply, first coefficients are multiplied with each other and the sub-radical amounts each other, placing the latter product under the radical sign common and the result is simplified.
But if you want to divide, then the coefficients are divided among themselves and sub-radical amounts each other, placing the latter quotient under the radical common and the result is simplified.
Recently Updated Pages
Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE
The highest possible oxidation states of Uranium and class 11 chemistry CBSE
Find the value of x if the mode of the following data class 11 maths CBSE
Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE
A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE
Statement I Reactivity of aluminium decreases when class 11 chemistry CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Difference Between Plant Cell and Animal Cell
What is pollution? How many types of pollution? Define it
What is the color of ferrous sulphate crystals? How does this color change after heating? Name the products formed on strongly heating ferrous sulphate crystals. What type of chemical reaction occurs in this type of change.