Question

pls answer the question quickly

Answer 1

Answer:

$\frac{ \sqrt{3} - 1 }{ \sqrt{3} + 1} \times \frac{ \sqrt{3} - 1}{ \sqrt{3} - 1} = \frac{( \sqrt{3} + 1 {)}^{2} }{ { \sqrt{3} }^{2} - {1}^{2} } \\ = \frac{3 + 1 + 2 \sqrt{3} }{3 - 1} = \frac{4 + 2 \sqrt{3} }{2} \\ = \frac{2(2 + \sqrt{3} )}{2} = 2 + \sqrt{3}$

now,

$a + b \sqrt{3} = 2 + \sqrt{3} \\ a = 2 \: and \: b = 1$