Question

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Answer 1

Answer:

a=4 and b=-1

Step-by-step explanation:

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given,

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

L.H.S,=

$= \frac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} - \sqrt{3} } \\ = \frac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} - \sqrt{3} } \times \frac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} + \sqrt{3} } \\ = \frac{( \sqrt{5} \times \sqrt{5} ) + ( \sqrt{5} \times \sqrt{3} ) + ( \sqrt{3} \times \sqrt{5} ) + ( \sqrt{3} \times \sqrt{3}) }{( \sqrt{5} - \sqrt{3} )( \sqrt{5} + \sqrt{3} } \\ = \frac{5 + \sqrt{15} + \sqrt{15} + 3 }{( { \sqrt{5} }^{2} - { \sqrt{3} }^{2} )} \\ = \frac{8 + 2 \sqrt{15} }{5 - 3} \\ = \frac{8 + 2 \sqrt{15} }{2} \\ = \frac{8}{2} + \frac{2 \sqrt{15} }{2} \\ = 4 + \sqrt{15}$

comparing LHS with RHS we get,

a=4 and b=-1