Question

If x =5+2√6, then Find ( x + 1/x)

Answer 1

Given :-  

• x = 5 + 2√6

To Find :-

•  Find ( x + 1/x)

Solution :-

$\sf : \; \implies 5 + 2 \sqrt{6} + \dfrac{1}{5+ 2 \sqrt{6}}$

$\sf : \; \implies 5 + 2 \sqrt{6} + \dfrac{1}{5 +2 \sqrt{6}} \times \dfrac{5- 2 \sqrt{6}}{ 5- 2 \sqrt{6}}$

$\sf : \; \implies 5 + 2 \sqrt{6} + \dfrac{ 1 \times ( 5- 2 \sqrt{6})}{ ( 5+ 2 \sqrt{6}) \times ( 5- 2 \sqrt{6} )}$

$\sf : \; \implies 5 + 2 \sqrt{6} + \dfrac{ 5- 2 \sqrt{6}}{ (5)^{2}- ( 2 \sqrt{6})^{2}}$

$\sf : \; \implies 5 + 2 \sqrt{6} + \dfrac{5- 2 \sqrt{6}}{ 25 -4 \times ( \sqrt{6})^{2}}$

$\sf : \; \implies 5 + 2 \sqrt{6} + \dfrac{5- 2 \sqrt{6}}{25-4 \times 6}$

$\sf : \; \implies 5 + 2 \sqrt{6} + \dfrac{5-2 \sqrt{6}}{25-24}$

$\sf : \; \implies 5 + 2 \sqrt{6} + \dfrac{5-2 \sqrt{6}}{1}$

$\sf : \; \implies 5 +2 \sqrt{6} + 5- 2 \sqrt{6}$

$\sf : \; \implies 5 + 5$

$\boxed{\bf{ \star \;\; 10 }}$

Answer 2

Given :-

• If value of x is 5+2√6

To Find :-

• Value of ( x + 1/x )

Solution :-

$\sf \dashrightarrow x + \dfrac{1}{x}$

$\sf \dashrightarrow \bigg\{ 5 + 2\sqrt{6} \bigg\} + \bigg\{ \dfrac{1}{5 + 2\sqrt{6}} \bigg\}$

$\sf \dashrightarrow \bigg\{ 5 + 2\sqrt{6} \bigg\} + \bigg\{ \dfrac{1}{5 + 2\sqrt{6}} \times \dfrac{5 - 2\sqrt{6}}{5-2\sqrt{6}} \bigg\}$

$\sf \dashrightarrow \bigg\{ 5 + 2\sqrt{6} \bigg\} + \bigg\{ \dfrac{1( 5 - 2\sqrt{6})}{(5)^{2} - (2)^{2} \times (\sqrt{6})^{2} } \bigg\}$

$\sf \dashrightarrow \bigg\{ 5 + 2\sqrt{6} \bigg\} + \bigg\{ \dfrac{1( 5 - 2\sqrt{6})}{25 - 4 \times 6 } \bigg\}$

$\sf \dashrightarrow \bigg\{ 5 + 2\sqrt{6} \bigg\} + \bigg\{ \dfrac{1( 5 - 2\sqrt{6})}{25 - 24 } \bigg\}$

$\sf \dashrightarrow \bigg\{ 5 + 2\sqrt{6} \bigg\} + \bigg\{ \dfrac{1( 5 - 2\sqrt{6})}{1} \bigg\}$

$\sf \dashrightarrow 5 + 2\sqrt{6} + 5 - 2 \sqrt{6}$

$\begin{gathered}{ \underline{ \red{ \boxed{ \sf{ x + \dfrac{1}{x}= 10}}}}} \end{gathered}$