Answer
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Hint: In this type of question we try to remove the radical by rationalizing the fraction. So we have to multiply and divide the given fraction with a number to make the denominator, which is a radicand (a number under a radical sign), a rational number.
Complete step-by-step solution:
The expression given in the question is $\dfrac{6}{{\sqrt 2 }}$
We simply have to ‘Rationalize’ this number.
Rationalization (Root Rationalization): A process in which radicals of an algebraic fraction are removed by making the denominator rational.
In simple terms, the denominator should not be an irrational number and to make it rational we do this process known as rationalization.
To make the denominator rational in this problem, we have to multiply $\sqrt 2 $by itself so that the radical gets cancelled out i.e.
$\sqrt 2 \times \sqrt 2 = {(\sqrt 2 )^2} = 2$
And here 2 is a rational number.
Applying this process to the given expression
$\dfrac{6}{{\sqrt 2 }} \times \dfrac{{\sqrt 2 }}{{\sqrt 2 }} = \dfrac{{6\sqrt 2 }}{2}$
Hence, this result is the simplified form of the given expression.
Note:There is no problem with an irrational denominator but to make the fraction in simpler form, rationalization is required. When the denominator has more than one element for example, $\sqrt 2 + 1,\;\sqrt 3 - 1,\sqrt 5 + \sqrt 2 $etc. then there is another way to make it rational. To do so, we multiply both the top and bottom of the fraction by Conjugate (Two similar terms with opposite sign between its elements) of the denominator e.g. $x + 1\;{\text{and}}\;x - 1$are conjugates.
So to simplify an expression like $\dfrac{1}{{5 - \sqrt 3 }}$ we proceed like this:
$\Rightarrow \dfrac{1}{{5 - \sqrt 3 }} \times \dfrac{{5 + \sqrt 3 }}{{5 + \sqrt 3 }}$
Here $5 + \sqrt 3 $is the conjugate of$5 - \sqrt 3 $
$\Rightarrow \dfrac{{5 + \sqrt 3 }}{{(5 - \sqrt 3 ) \times (5 + \sqrt 3 )}}$
Here we use the formula in the denominator will be
$(a - b) \times (a + b) = {a^2} - {b^2}$
So we have
$\Rightarrow \dfrac{{5 + \sqrt 3 }}{{25 - 3}} = \dfrac{{5 + \sqrt 3 }}{{22}}$
This is the simplified form of $\dfrac{1}{{5 - \sqrt 3 }}$
Complete step-by-step solution:
The expression given in the question is $\dfrac{6}{{\sqrt 2 }}$
We simply have to ‘Rationalize’ this number.
Rationalization (Root Rationalization): A process in which radicals of an algebraic fraction are removed by making the denominator rational.
In simple terms, the denominator should not be an irrational number and to make it rational we do this process known as rationalization.
To make the denominator rational in this problem, we have to multiply $\sqrt 2 $by itself so that the radical gets cancelled out i.e.
$\sqrt 2 \times \sqrt 2 = {(\sqrt 2 )^2} = 2$
And here 2 is a rational number.
Applying this process to the given expression
$\dfrac{6}{{\sqrt 2 }} \times \dfrac{{\sqrt 2 }}{{\sqrt 2 }} = \dfrac{{6\sqrt 2 }}{2}$
Hence, this result is the simplified form of the given expression.
Note:There is no problem with an irrational denominator but to make the fraction in simpler form, rationalization is required. When the denominator has more than one element for example, $\sqrt 2 + 1,\;\sqrt 3 - 1,\sqrt 5 + \sqrt 2 $etc. then there is another way to make it rational. To do so, we multiply both the top and bottom of the fraction by Conjugate (Two similar terms with opposite sign between its elements) of the denominator e.g. $x + 1\;{\text{and}}\;x - 1$are conjugates.
So to simplify an expression like $\dfrac{1}{{5 - \sqrt 3 }}$ we proceed like this:
$\Rightarrow \dfrac{1}{{5 - \sqrt 3 }} \times \dfrac{{5 + \sqrt 3 }}{{5 + \sqrt 3 }}$
Here $5 + \sqrt 3 $is the conjugate of$5 - \sqrt 3 $
$\Rightarrow \dfrac{{5 + \sqrt 3 }}{{(5 - \sqrt 3 ) \times (5 + \sqrt 3 )}}$
Here we use the formula in the denominator will be
$(a - b) \times (a + b) = {a^2} - {b^2}$
So we have
$\Rightarrow \dfrac{{5 + \sqrt 3 }}{{25 - 3}} = \dfrac{{5 + \sqrt 3 }}{{22}}$
This is the simplified form of $\dfrac{1}{{5 - \sqrt 3 }}$
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