Question

Rationalise the denominator of root five by root three +root 2

Answer 1

Hi there !

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Given :

√5 / √3 + √2

Solution :

√5 / √3 + √2

multiply numerator and denominator by √3 - √2.

As follows :)


√5 / √3 + √2 × √3 - √2 / √3 - √2

√5 ( √3 - √2) / (√3)² - (√2)²

√15 - √10 / 2 - 1

√15 - √10

Hence,

On rationalising it's denominator, we get ;

=> √15 - √10

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Thanks for the question !

Answer 2

$\huge \red{hello}$
$\pink{given}$
$= > \frac{ \sqrt{5} }{ \sqrt{3 + \sqrt{2} } }$
$\pink{now}$
$\pink{multiply \: by \: \sqrt{3 - \sqrt{2} \: \: in} }$
$\pink{numerator \: and \: denominator}$
$= > \: \frac{ \sqrt{5 \times \sqrt{3 - \sqrt{2} } } }{ \sqrt{3 + \sqrt{2 \times \sqrt{3 - \sqrt{2} } } } }$
$= > \frac{ \sqrt{15 - \sqrt{10} } }{ \sqrt{ {3}^{2} - \sqrt{ {2}^{2} } } }$

Reason Formula is
(a+b)(a-b)

$= > \frac{ \sqrt{15 - \sqrt{10} } }{3 - 2}$
$= > \frac{ \sqrt{15 - \sqrt{10} } }{1}$
$= > \sqrt{15 - \sqrt{10} }$

hence,
The Rationalised form is
$\sqrt{15 - \sqrt{10} }$
$\huge \green{thank \: you}$