Question

hey...
....
.....
plz help

Answer 1

Heya dear,
Here is the answer you were looking for:

Given,
√2 = 1.414

$\sqrt{ \frac{ \sqrt{2 - 1} }{ \sqrt{2 + 1} } } \\ \\ = \sqrt{ \frac{ \sqrt{2} - \sqrt{1} }{ \sqrt{2} + \sqrt{1} } }$

On rationalizing the denominator we get,

$= \sqrt{ \frac{ \sqrt{2} - \sqrt{1} }{ \sqrt{2} + \sqrt{1} } \times \frac{ \sqrt{2} - \sqrt{1} }{ \sqrt{2} - \sqrt{1} } }$

Using the identity:

${(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy \\ (x + y)(x - y) = {x}^{2} - {y}^{2}$

$= \sqrt{ \frac{ {( \sqrt{2}) }^{2} + {( \sqrt{1}) }^{2} - 2( \sqrt{2})( \sqrt{1} )}{ {( \sqrt{2}) }^{2} - {( \sqrt{1} )}^{2} } } \\ \\ = \sqrt{ \frac{2 + 1 - 2 \sqrt{2} }{2 - 1} } \\ \\ = \sqrt{3 - 2 \sqrt{2} }$


We can write it as

$\sqrt{ { (\sqrt{2} )}^{2} + {( \sqrt{1}) }^{2} - 2( \sqrt{2})( \sqrt{1}) } \\ \\ = \sqrt{ {( \sqrt{2} + \sqrt{1} )}^{2} } \\ \\ = \sqrt{2} + \sqrt{1} \\ \\ = \sqrt{2} + {( \sqrt{1} )}^{2} \\ \\ = \sqrt{2} + 1 \\ \\ = 1.414 + 1 \\ \\ = 2.414$


Hope this helps!!

If you have any doubt regarding to my answer, feel free to ask in the comment section or inbox me if needed.

@Mahak24

Thanks...
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