Question

if√3+√2/√3_√2=a+b√6 then b=​

Answer 1

$\color{palevioletred} \underline {\underline{\mathfrak{ {Given:}}}} \\ </h3><p> \small\rm{\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3} - \sqrt{2} } = {a + b \sqrt{6}} } \\ \\ </p><p>\color{palevioletred} \underline {\underline{\mathfrak{ {To \: Find : }}}} \\ \rm{values \: of \: \: a \: \: and \: \: b} \\ \\ \color{palevioletred} \underline {\underline{\mathfrak{ {Solution:}}}} \: \\ \rm{rationalise \: the \: denominator \: of \: \small\rm{\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3} - \sqrt{2}}}}$

$\implies\small\rm{\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3} - \sqrt{2}} \times \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } = a + b \sqrt{6} } \\ \\ \implies{ \rm{ \dfrac{ {( \sqrt{3} + \sqrt{2} ) }^{2} }{ {( \sqrt{3} )}^{2} - {( \sqrt{2} )}^{2} } = a + b \sqrt{6} }} \\ \\ \implies{ \rm{ \frac{ {( \sqrt{3} )}^{2} + {( \sqrt{2} )}^{2} }{3 - 2} } = a + b \sqrt{6} } \\ \\ \implies\small\rm{\dfrac{{{{3} }^{} +{2} }^{} + 2\sqrt{6} }{{{}}^{} { {1}} } = a + b \sqrt{6} } \\ \\ \implies \: \rm \: 5 + 2 \sqrt{6} = a + b \sqrt{6}$

On comparing both sides we get,

a = 5

b = 2