Question

if a and b are any two rational numbers and if (√3+√2)/(√3-√2) = a+b√6 ,then find a and b​

Answer 1

Answer:

• a = 5
• b = 2

Explanation:

Given,

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

Taking L. H. S.

$\longrightarrow \: \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$

Rationalising the denominator,

$\longrightarrow \: \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} }$

$\longrightarrow \: \dfrac{( \sqrt{3} + \sqrt{2} )( \sqrt{3} + \sqrt{2} )}{( \sqrt{3} - \sqrt{2} )( \sqrt{3} + \sqrt{2} )}$

${{ \longrightarrow \: \dfrac{ \sqrt{3} (\sqrt{3} + \sqrt{2} ) + \sqrt{2} ( \sqrt{3} + \sqrt{2}) }{ {( \sqrt{3} )}^{2} - {( \sqrt{2} )}^{2} } } \: \: \: \: \: \: \: \{ \because \: (a - b)(a + b) = {a}^{2} - {b}^{2} \}}$

$\longrightarrow \: \dfrac{ {( \sqrt{3}) }^{2} + \sqrt{6} + \sqrt{6} + {( \sqrt{2} )}^{2} }{3 - 2}$

$\longrightarrow \: \dfrac{3 + 2 \sqrt{6} + 2 }{1}$

$\longrightarrow \: 5 + 2 \sqrt{6}$

On comapring with R. H. S. i.e., a + b√6

$\implies \: 5 + 2 \sqrt{6} = a + b \sqrt{6}$

$\implies \: a = 5 \: { \rm \: and} \: b = 2$