Question

Root3-1/root3+1=a+b root3

Answer 1

By Rationalising the denominator, we have

$\large\bold{\frac{ \sqrt{3} - 1}{ \sqrt{3} + 1} = a + b \sqrt{3}}$

$\large \frac{ \sqrt{3} - 1}{ \sqrt{3} + 1 } = \frac{ \sqrt{3} - 1}{ \sqrt{3} + 1 } \times \frac{ \sqrt{3 } - 1}{ \sqrt{3} - 1 }$

$\large= \frac{( \sqrt{3} - 1)( \sqrt{3} - 1) }{( \sqrt{3} + 1)( \sqrt{3} - 1) }$

$\large= \frac{( \sqrt{3} - 1)^{2} }{ (\sqrt{3} ) ^{2} - {1}^{2} }$

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

$\large= \frac{3 + 1 - 2 \sqrt{3} }{3 - 1}$

$\large = \frac{4 - 2 \sqrt{3} }{2}$

$\large = 2 - \sqrt{3}$

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

$\large⇒2 - \sqrt{3} = a + b \sqrt{3}$

$\large⇒a + b \sqrt{3} = 2 + ( - 1) \sqrt{3}$

$\large\fbox\bold{ a = 2 \: and \: b = - 1}$

[ By equating rational and irrational parts ]