Question

evaluate √5+√2 by √5-√2. given that √10=3.162

Answer 1

Identities used :

${(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \\ (a + b)(a - b) = {a}^{2} - {b}^{2}$

Given,
√10 = 3.162

Now,

$\frac{ \sqrt{5} + \sqrt{2} }{ \sqrt{5} - \sqrt{2} }$

On rationalizing the denominator we get,

$= \frac{ \sqrt{5} + \sqrt{2} }{ \sqrt{5} - \sqrt{2} } \times \frac{ \sqrt{5} + \sqrt{2} }{ \sqrt{5} + \sqrt{2} } \\ \\ = \frac{ {( \sqrt{5} )}^{2} + {( \sqrt{2} )}^{2} + 2( \sqrt{5} )( \sqrt{2}) }{ {( \sqrt{5}) }^{2} - {( \sqrt{2} )}^{2} } \\ \\ = \frac{5 + 2 + 2\sqrt{10} }{5 - 2} \\ \\ = \frac{7 +2 \sqrt{10} }{3} \\ \\ = \frac{7 + 2 \times 3.162}{3} \\ \\ = \frac{7 + 6.324}{3} \\ \\ = \frac{13.324}{3} \\ \\ = 4.441 \: (approx)$

Answer 2

Heya friend !!!!

•°• Here's your answer •°•

$que = \frac{ \sqrt{5} + \sqrt{2} }{ \sqrt{5} - \sqrt{2} } \\ \\ = \frac{ \sqrt{5} + \sqrt{2} }{ \sqrt{5} - \sqrt{2} } \times \frac{ \sqrt{5} + \sqrt{2} }{ \sqrt{5} + \sqrt{2} } \\ \\ identities \: \: used \: \: \\ \\ = (a + b)(a + b) = {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \\ = (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = \frac{ {( \sqrt{5} + \sqrt{2} )}^{2} }{ { (\sqrt{5}) }^{2} - { (\sqrt{2} )}^{2} } \\ \\ = \frac{ {( \sqrt{5} )}^{2} + {( \sqrt{2} )}^{2} + 2 \times \sqrt{5} \times \sqrt{2} }{ \sqrt{5 \times 5} - \sqrt{2 \times 2} } \\ \\ = \frac{5 + 2 + 2 \sqrt{10} }{5 - 2} \\ \\ = put \: \: \sqrt{10} = 3.162 \\ \\ = \frac{7 + 2 \times 3.162}{3} \\ \\ = \frac{7 + 6.324}{3} \\ \\ = \frac{13.324}{3}$

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