Question

in this question a nd b value​

Answer 1

$\large\underline\blue{\bold{Given \:  Question :-  }}$

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

find the value of a and b.

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${\large{\bold{\rm{\underline{Full \; Solution}}}}}$

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

☆ On rationalizing the denominator, we get

$\tt \:  \longrightarrow \: \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } = a \: + \: b \: \sqrt{6}$

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

$\tt \:  \longrightarrow \: \dfrac{ {( \sqrt{3}) }^{2} + {( \sqrt{2} )}^{2} + 2 \times \sqrt{2} \times \sqrt{3} }{3 - 2} = a \: + \: b \sqrt{6}$

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

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

$\tt\implies \:a \: = 5 \: and \: b \: = \: 2$

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