Question

$\frac{ \sqrt{5} + \sqrt{3} }{2 \sqrt{5} - 3 \sqrt{3} } = a - b \sqrt{15}$

Answer 1

Using the identity

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

$\frac{ \sqrt{5} + \sqrt{3} }{2 \sqrt{5} - 3 \sqrt{3} } = a - b \sqrt{15}$

L.H.S,

On rationalizing the denominator we get,

$= \frac{ \sqrt{5} + \sqrt{3} }{2 \sqrt{5} - 3 \sqrt{3} } \times \frac{ 2\sqrt{5} + 3 \sqrt{3} }{2 \sqrt{5} + 3 \sqrt{3} } \\ \\ = \frac{ \sqrt{5}(2 \sqrt{5} + 3 \sqrt{3}) + \sqrt{3}(2 \sqrt{5} + 3 \sqrt{3} ) }{ {(2 \sqrt{5} )}^{2} - {(3\sqrt{3}) }^{2} } \\ \\ = \frac{10 + 3 \sqrt{15} + 2 \sqrt{15} + 9 }{20 - 27} \\ \\ = \frac{19 + 5 \sqrt{15} }{ - 7} \\ \\ = \frac{ - 19 - 5 \sqrt{15} }{7}$

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

a = - 19/7

b = 5/7