Question

rationalize..........................?

Answer 1

Rationalizing the denominator!!

So, We need to know What is Rationalisation?
$\bf → \: So, \: The \: process \: in \: which \: we \\ \bf rewrite \: the \: Denominator \: in \\ \bf rational \: form$

Now, Let's solve the given Problems :

$\bf (i) \: \frac{1}{ \sqrt{7} }$

→ So, To rationalise the denominator, we will multiply the given denominator to both (numerator and denominator)

$= \frac{1}{ \sqrt{7} } \times \frac{ \sqrt{7} }{ \sqrt{7} } \\ \\ = \frac{ \sqrt{7} }{ {( \sqrt{7} )}^{2} } \\ \\ = \frac{ \sqrt{7} }{7}$

$\bf (ii) \: \frac{1}{ \sqrt{7} - \sqrt{6} }$

→ Now, we'll have to multiply something which would ultimately give a rational number in the denominator.

So, let's try multiplying √7 + √6 to √7 - √6 which is same as (a + b)(a - b) which is equal to $\bf {a}^2 - {b}^2$

Here we go,

$= \frac{1}{ \sqrt{7} - \sqrt{6} } \times \frac{ \sqrt{7} + \sqrt{6} }{ \sqrt{7} + \sqrt{6} } \\ \\ = \frac{ \sqrt{7} + \sqrt{6} }{ {( \sqrt{7}) }^{2} - {( \sqrt{6}) }^{2} } \\ \\ = \frac{ \sqrt{7} + \sqrt{6} }{7 - 6} \\ \\ = \frac{ \sqrt{7} + \sqrt{6} }{1} \: or \: \sqrt{7} + \sqrt{6}$

Similarly,

$\bf (iii) \: \frac{1}{ \sqrt{5} + \sqrt{2} } \\ \\ = \frac{1}{ \sqrt{5} + \sqrt{2} } \times \frac{ \sqrt{5} - \sqrt{2} }{ \sqrt{5} - \sqrt{2} } \\ \\ = \frac{ \sqrt{5} - \sqrt{2} }{ {( \sqrt{5}) }^{2} - {( \sqrt{2} )}^{2} } \\ \\ = \frac{ \sqrt{5} - \sqrt{2} }{5 - 2} \\ \\ = \frac{ \sqrt{5} - \sqrt{2} }{3}$

$\bf (iv) \frac{1}{ \sqrt{7} - 2 } \\ \\ = \frac{1}{ \sqrt{7} - 2} \times \frac{ \sqrt{7} + 2}{ \sqrt{7} + 2 } \\ \\ = \frac{ \sqrt{7} + 2}{ {( \sqrt{7} )}^{2} - {(2)}^{2} } \\ \\ = \frac{ \sqrt{7} + 2 }{7 - 4} \\ \\ = \frac{ \sqrt{7} + 2}{3}$

If any doubt, please ask ;)