Question

If a,b are rational numbers find value of a,b Root3+root2 by root3-root2=a+b root6

Answer 1

Identities used :

${(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \\ (a + b)(a - b) = {a}^{2} - {b}^{2}$

Now,

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

L.H.S,

On rationalizing the denominator we get,

$= \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ = \frac{ {( \sqrt{3} )}^{2} + {( \sqrt{2} )}^{2} + 2( \sqrt{3})( \sqrt{2} ) }{ {( \sqrt{3} )}^{2} - {( \sqrt{2}) }^{2} } \\ \\ = \frac{3 + 2 + 2 \sqrt{6} }{3 - 2} \\ \\ = 5 + 2 \sqrt{6}$

On comparing L.H.S and R.H.S we get,

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