Question

find teh value of a and b of √3-1÷√3+1=a+b√3​

Answer 1

Answer:

a = 2, b = -1

Step-by-step explanation:

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

Therefore, a + b√3 = 2 - √3

a + b√3 = 2 + (-1)√3

a = 2, b = -1