Question

if a =2+√3 what would be the value of a+1/a​

Answer 1

Given :-

$a = 2 + \sqrt{3}$

To Find :-

(a+1)/a = ?

Solution :-

$= > \frac{a + 1}{a}$

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

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

》 At this stage we have to use determenent to get value easily.

》 if deteminent is 》"x+y" we should use the appropriate as "x-y".

》Then only we can remove that root easily.

$= > \frac{3 + \sqrt{3} }{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} }$

$= > \frac{(3 + \sqrt{3})(2 - \sqrt{3}) }{(2 + \sqrt{3} )(2 - \sqrt{3}) }$

$= > \frac{6 - 3 \sqrt{3} + 2 \sqrt{3} - ( \sqrt{3} {}^{2} ) }{(2) {}^{2} - ( \sqrt{3}) {}^{2} }$

$= > \frac{6 - \sqrt{3} - 3 }{4 - 3}$

$= > \frac{3 - \sqrt{3} }{1}$

$= > 3 - \sqrt{3}$

Formulas or rules :-

(a+b)(a-b) = (a^2 - b^2)

Answer 2

Given,

$\boxed{ \boxed{ \rm{a = 2 + \sqrt{3} } \: \: } \: \: \: }$

To find :- $\boxed{ \rm{a + \frac{ 1}{a} }\: \: \: \: }$

$\underline {\huge \dag \: \: \text{ \red{S}O\red{L}U\red{T}I\red{O}N :}}$

Here,

$\boxed{ \rm{}a = 2 + \sqrt{3} \: \: \: }$

$\boxed{ \rm \frac{1}{a} = \frac{1}{2 + \sqrt{3} } \: \: }$

Now, Simplifying the value 1/a ,

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

Thus,

$\huge = > \: \: \rm{}a + \frac{1}{a} \\ \\ \bf= ( 2 + \sqrt{3} ) + (2 - \sqrt{3}) \\ \bf = \: 2 + \cancel{ \sqrt{3} }+ 2 - \cancel{\sqrt{3}} \: \: \: \: \: \: \: = \red4$

Hence,

$\boxed {\boxed {\boxed { \large \boxed { \: \: \rm{a + \frac{1}{a} = 4} \: \: \: \: \: }}}}$