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Answer 1

Question:-

If

$\sf\dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } = a + b \sqrt{6}$

then (a , b) =

Answer:-

Given:

$\:\sf\dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } = a + b \sqrt{6}$

Multiplying numerator and denominator by √3 + √2 & √3 + √2 in LHS we get,

$\implies\sf\dfrac{ (\sqrt{3} + \sqrt{2} )( \sqrt{3} + \sqrt{2} ) }{ (\sqrt{3} - \sqrt{2} )( \sqrt{3} + \sqrt{2} )} = a + b \sqrt{6}$

using -

• (a + b)(a + b) = (a + b)² = a² + b² + 2ab

• (a - b)(a + b) = a² - b²

we get,

$\implies \sf \: \dfrac{( \sqrt{3}) ^{2} + {( \sqrt{2} )}^{2} + 2 \times \sqrt{3} \times \sqrt{2} }{( \sqrt{3} ) ^{2} - {( \sqrt{2}) }^{2} } = a + b \sqrt{6} \\\\ \\ \implies \sf \: \dfrac{3 + 2 + 2 \times \sqrt{6} }{( \sqrt{3} ) ^{2} - {( \sqrt{2}) }^{2} } = a + b \sqrt{6} \\\\ \\ \implies \sf \dfrac{5 + 2 \sqrt{6} } {3 - 2} = a + b \sqrt{6} \\ \\\\ \implies \boxed{\sf \: a + b \sqrt{6} = 5 + 2 \sqrt{6} }$

In comparison of both sides we get,

• a = 5
• b = 2.

∴ (a , b) = (5 , 2)