If a=2+√3, then a+1/a=?

1.	If a=2+√3, then a+1/a=?
Answer» align="ABSMIDDLE" alt="\huge\blue{\tt{Question:-}}" CLASS="latex-formula" id="TexFormula1" src="https://tex.z-dn.net/?f=%5Chuge%5Cblue%7B%5Ctt%7BQuestion%3A-%7D%7D" TITLE="\huge\blue{\tt{Question:-}}"> $If \: a = (2 + \sqrt{3} ) , \: then \: a + \frac{1}{a} = ?$ $\huge\blue{\tt{Solution:-}}$ $a + \frac{1}{a} = (2 + \sqrt{3} ) + \frac{1}{(2 + \sqrt{3} )}$ $solving \: \frac{1}{a}$ Using rationalisation, $\frac{1}{(2 + \sqrt{3} )} \times \frac{(2 - \sqrt{3} )}{(2 - \sqrt{3} )}$ $Since, (a+b)(a-b) = {a}^{2} - {b}^{2}$ Therefore, $= > \frac{(2 - \sqrt{3}) }{ {(2)}^{2} - {( \sqrt{3} )}^{2} }$ $= > \frac{(2 - \sqrt{3}) }{<klux>4</klux> - 3}$ $= > \frac{(2 - \sqrt{3}) }{1} = (2 - \sqrt{3} )$ $Now, a + \frac{1}{a} = (2 + \sqrt{3} ) + (2 - \sqrt{3} )$ => 2 + 2 => 4. Hence, 4 is the required answer.

Answer»

align="ABSMIDDLE" alt="\huge\blue{\tt{Question:-}}" CLASS="latex-formula" id="TexFormula1" src="https://tex.z-dn.net/?f=%5Chuge%5Cblue%7B%5Ctt%7BQuestion%3A-%7D%7D" TITLE="\huge\blue{\tt{Question:-}}">

$If \: a = (2 + \sqrt{3} ) , \: then \: a + \frac{1}{a} = ?$