Question

Write the simplest rationalization factor in the following surd: $\sqrt {32} $
A. $\sqrt 2 $
B. $\sqrt 7 $
C. $\sqrt 5 $
D. $\sqrt 3 $

Answer 1

Hint:
In the given question we have to find the simplest (least value) rationalization factor of $\sqrt {32} $.
So, first express $\sqrt {32} $ in the simplest form.
By prime factorization of 32 you will get $\sqrt {32} = \sqrt {2 \times 2 \times 2 \times 2 \times 2} = 4\sqrt 2 $.
Now find the simplest irrational number which on multiplying with $4\sqrt 2 $ will give a rational value, that number will be the rationalization factor of $\sqrt {32} $.

Complete step by step solution:
A surd is an unresolved root or sum of roots. It is an irrational number which cannot be represented in the form of fractions of whole numbers.
Now, the process of multiplying a surd with another surd, so that their multiplication gives a rational value is called rationalization. And both the surds are called rationalization factors of each other.
For example: $\sqrt 2 $ multiplied by $\sqrt 2 $ gives 2, which is a rational number. Hence both are rationalization factors of each other.
Here, in the given question we have to find the simplest (least value) rationalization factor of $\sqrt {32} $.
By prime factorization of 32 we can write that:$\sqrt {32} \times \sqrt 2 = 4\sqrt 2 \times \sqrt 2 = 4 \times 2 = 8$
$\sqrt {32} = \sqrt {\underline {2 \times 2} \times \underline {2 \times 2} \times 2} $
The two pairs of $2 \times 2$ can be taken out of the root as 2, because the square root of 4 is 2.
$\therefore \sqrt {32} = 2 \times 2\sqrt 2 $
$ \Rightarrow \sqrt {32} = 4\sqrt 2 $
So, we can easily see that when we multiply $\sqrt {32} $ by $\sqrt 2 $ it will give a rational value which is 8.

Hence, $\sqrt 2 $ is the simplest rationalization factor of $\sqrt {32} $.

Note:
We generally need to rationalize the denominator of a fraction which has an irrational number as its denominator.
For rationalizing the denominator, we just multiply both the numerator and denominator by the same irrational number as the denominator of that fraction.
$\dfrac{a}{{\sqrt b }} = \dfrac{a}{{\sqrt b }} \times \dfrac{{\sqrt b }}{{\sqrt b }} = \dfrac{{a \times \sqrt b }}{{\sqrt b \times \sqrt b }} = \dfrac{{a\sqrt b }}{b}$