Question

Find the value of a and b.

$if = \frac{ \sqrt{3 - 1} }{ \sqrt{3 + 1} } = a + b \sqrt{3}$
​

Answer 1

Answer:

$a = 2 \: and \: b = ( - 1)$

Step-by-step explanation:

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