Question

If x=√7+√2/√7-√2 then find the value of x^2+1/x^2

Answer 1

ANSWER :–

$\\ \to { \boxed{ \bold{ {x}^{2} + \dfrac{1}{ {x}^{2} } =\frac{ 274 }{25} }}}$

EXPLANATION :–

GIVEN :–

▪︎${ \bold{x = \dfrac{ \sqrt{7} + \sqrt{2} }{ \sqrt{7} - \sqrt{2} } }}$

TO FIND :–

▪︎ ${ \bold{Value \: \: of \: \: \to {x}^{2} + \dfrac{1}{ {x}^{2} } }}$

SOLUTION :–

▪︎ We know that –

$\\ \implies { \boxed{\bold{(a + b)^{2} = {a}^{2} + {b}^{2} + 2ab }}}$

▪︎ So that –

$\\ \implies {\bold{(x + \frac{1}{x} )^{2} = {x}^{2} + \dfrac{1}{ {x}^{2} } + 2(x)( \frac{1}{x} ) }}$

$\\ \implies {\bold{(x + \frac{1}{x} )^{2} = {x}^{2} + \dfrac{1}{ {x}^{2} } + 2}}$

$\\ \implies {\bold{ {x}^{2} + \dfrac{1}{ {x}^{2} } = (x + \frac{1}{x} )^{2} - 2 }}$

▪︎ Now put the values –

$\\ \implies { \bold{ {x}^{2} + \dfrac{1}{ {x}^{2} } = (\dfrac{ \sqrt{7} + \sqrt{2} }{ \sqrt{7} - \sqrt{2} } + \dfrac{ \sqrt{7} - \sqrt{2} }{ \sqrt{7} + \sqrt{2} } )^{2} - 2 }}$

$\\ \implies { \bold{ {x}^{2} + \dfrac{1}{ {x}^{2} } =[ \frac{ {( \sqrt{7} + \sqrt{2} )}^{2} + {( \sqrt{7} - \sqrt{2} )}^{2} }{ {( \sqrt{7}) }^{2} - {( \sqrt{2}) }^{2} } ]^{2} - 2 }}$

$\\ \implies { \bold{ {x}^{2} + \dfrac{1}{ {x}^{2} } =[ \frac{ 7 + 2 + 2 \sqrt{14} + 7 + 2 - 2 \sqrt{14} }{ 7 - 2 } ]^{2} - 2 }}$

$\\ \implies { \bold{ {x}^{2} + \dfrac{1}{ {x}^{2} } =(\frac{ 18 }{5} )^{2} - 2 }}$

$\\ \implies { \bold{ {x}^{2} + \dfrac{1}{ {x}^{2} } =(\dfrac{ 324 }{25} )- 2 }}$

$\\ \implies { \boxed{ \bold{ {x}^{2} + \dfrac{1}{ {x}^{2} } =\frac{ 274 }{25} }}}$