Question

If x=5-root21/2 , find x+1/x​

Answer 1

Question :

If $\sf{x = \frac{ 5 - \sqrt{21}}{2} } \: then \: find \: the \: value \: of \: x + \frac{1}{x}$

Answer :

Given :

• $\sf{x = \frac{ 5 - \sqrt{21}}{2} }$

To find :

• $\sf x + \frac{1}{x}$

According to the question,

$\implies\sf{x = \frac{5 - \sqrt{21} }{2} \times \frac{2}{2} }$

$\implies \sf{ \frac{2(5 - \sqrt{21}) }{4} }$

$\implies \sf{ \frac{10 - 2 \sqrt{21} }{4} }$

$\sf \implies{ \frac{5 - \sqrt{21} }{2} } \green \bigstar$

$\sf \implies{ \frac{1}{x} = \frac{2}{5 - \sqrt{21} } \times \frac{5 + \sqrt{21} }{5 + \sqrt{21} } }$

$\sf \implies{ \frac{2(5 + \sqrt{21}) }{25 - 21} }$

$\implies \sf{ \frac{10 + 2 \sqrt{21} }{4} }$

$\implies \sf{ \frac{5 + \sqrt{21} }{2} } \green \bigstar$

$\therefore \sf{x + \frac{1}{x} = \frac{5 - \sqrt{21} }{2} + \frac{5 + \sqrt{21} }{2} }$

$\sf \implies { \frac{5 - \cancel{\sqrt{21}} + 5 + \cancel{\sqrt{21} } }{2} }$

$\sf \implies{ \frac{10}{2} }$

$\implies { \green{ \boxed{\sf{5}}}}$