Question

find the value of a&b if √7-1/√7+1 -√7+1/√7-1=a+b√7​

Answer 1

Solution :-

$\dfrac{ \sqrt{7} - 1 }{ \sqrt{7} + 1 } - \dfrac{ \sqrt{7} + 1 }{ \sqrt{7} - 1 } = a + b \sqrt{7}$

Rationalising each term in LHS

$\bigg(\dfrac{ \sqrt{7} - 1 }{ \sqrt{7} + 1 } \times \dfrac{ \sqrt{7} - 1}{ \sqrt{7} - 1} \bigg)- \bigg(\dfrac{ \sqrt{7} + 1 }{ \sqrt{7} - 1 } \times \dfrac{ \sqrt{7} + 1}{ \sqrt{7} + 1 } \bigg) = a + b \sqrt{7}$

$\dfrac{ (\sqrt{7} - 1)^{2} }{ (\sqrt{7} )^{2} - {1}^{2} }- \dfrac{ (\sqrt{7} + 1)^{2} }{ (\sqrt{7})^{2} - {1}^{2} } = a + b \sqrt{7}$

[ Because (x + y)(x - y) = x² - y² ]

$\dfrac{ ( \sqrt{7})^{2} + {1}^{2} - 2( \sqrt{7})(1)}{7 - 1 }- \dfrac{( \sqrt{7})^{2} + {1}^{2} + 2( \sqrt{7})(1) }{7 - 1} = a + b \sqrt{7}$

[ Because (x + y)² = x² + y² + 2xy and (x - y)² = x² + y² - 2xy ]

$\dfrac{ 7+ 1 - 2 \sqrt{7}}{7 - 1 }- \dfrac{7+ 1 + 2\sqrt{7} }{7 - 1} = a + b \sqrt{7}$

$\dfrac{ 8- 2 \sqrt{7}}{6 }- \dfrac{8 + 2\sqrt{7} }{6} = a + b \sqrt{7}$

$\dfrac{ 8- 2 \sqrt{7} - (8 + 2 \sqrt{7} )}{6 }= a + b \sqrt{7}$

$\dfrac{ 8- 2 \sqrt{7} - 8 - 2 \sqrt{7} }{6 }= a + b \sqrt{7}$

$\dfrac{ - 4 \sqrt{7} }{6 }= a + b \sqrt{7}$

Comparing on both sides

$\implies a = 0 \qquad b \sqrt{7} = - \dfrac{4 \sqrt{7} }{6}$

$\implies a = 0 \qquad b = - \dfrac{4 \sqrt{7} }{6 \sqrt{7} }$

$\implies a = 0 \qquad b = - \dfrac{2}{3}$

Hence the value of a is 0 and the value of b is - 2/3.