Question

Find the value of A and B root 3 by 2 to 3 mai find the value of A and B root 3 + root 2 by root 3 minus root 2 is equal to a + b root 6

Answer 1

☆Given

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

☆ On rationalizing the denominator, we get

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

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

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

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

☆ On comparing both sides, we get

$\large\boxed{ \pink{ \bf \: a \: = 5 \: \: and \: \: b \: = 2}}$