Question

Plz solve............................

Answer 1

$a = 2 + \sqrt{3} \\ \\ = > \frac{1}{a} = \frac{1}{2 + \sqrt{3} }$

rationalising,

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


So,

1) a + 1/a

= 2 + √3 + 2 - √3

= 4


Answer :- 4


2) a² + 1/a²

=> (2 + √3)² + (2 - √3)²

= 4 + 3 + 4√3 + 4 + 3 - 4√3

= 14

Answer :- 14


3) a - 1/a

= 2 + √3 - (2 - √3)

= 2 + √3 - 2 + √3

= √3 + √3

= 2√3

4) √a - 1/√a

= √(2 + √3) - √(2 - √3)



Now, To find √(2 + √3)

=> Multiply 2 on both numerator and denominator

$\sqrt{2 + \sqrt{3} \times \frac{2}{2} } \\ \\ = > \sqrt{ \frac{4 + 2 \sqrt{3} }{2} } \\ \\ = > \sqrt{ \frac{( \sqrt{3) {}^{2} } + 1 {}^{2} + 2( \sqrt{3)1} }{2} } \\ \\ = > \sqrt{ \frac{( \sqrt{3} + 1) {}^{2} }{2} } \\ \\ = > \frac{ \sqrt{3} + 1 }{ \sqrt{2} } \\ \\ = > \frac{( \sqrt{3} + 1)( \sqrt{2)} }{ \sqrt{2} \times \sqrt{2} } \\ \\ = > \frac{ \sqrt{6 } + \sqrt{2} }{2}$


Similarly √(2 - √3) =
$\frac{ \sqrt{6} - \sqrt{2} }{2}$



Hence, √a - 1/√a

=
$\frac{ \sqrt{6} + \sqrt{2} }{2} - ( \frac{ \sqrt{6} - \sqrt{2} }{2} )$

$= > \frac{ \sqrt{6} + \sqrt{2} - \sqrt{6} + \sqrt{2} }{2} \\ \\ = > \frac{2 \sqrt{2} }{2} \\ \\ = \sqrt{2}$


Answer :- √2