Question

Find the value of a and b if i) √2+√3/3√2-2√3 = a+b√6

Answer 1

Hi !

Identity used :

$(x + y)(x - y) = {x}^{2} - {y}^{2}$

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

L.H.S

On rationalizing the denominator we get,

$= \frac{ \sqrt{2} + \sqrt{3} }{3 \sqrt{2} - 2 \sqrt{3} } \times \frac{3 \sqrt{2} + 2 \sqrt{3} }{3 \sqrt{2} + 2 \sqrt{3} } \\ \\ = \frac{ \sqrt{2} (3 \sqrt{2} + 2 \sqrt{3} ) + \sqrt{3} (3 \sqrt{2} + 2 \sqrt{3} )}{ {(3 \sqrt{2}) }^{2} - {(2 \sqrt{3}) }^{2} } \\ \\ = \frac{6 + 2 \sqrt{6} + 3 \sqrt{6} + 6}{18 - 12} \\ \\ = \frac{12 + 5 \sqrt{6} }{6} \\ \\ = 2 + \frac{5 \sqrt{6} }{6}$

On comparing we get,

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