Question

$x = \frac{ \sqrt{6} + \sqrt{7} }{ \sqrt{7} - \sqrt{6} } \: then \: find \: {(x + \frac{1}{x} )}^{2} =$

please help​

Answer 1

Given:

• $x = \dfrac{ \sqrt{6} + \sqrt{4} }{ \sqrt{6} + \sqrt{7} }$

Then, we can get the value of 1/x by finding value of the reciprocal of given value of x

$\dfrac{1}{ x } = \dfrac{1}{ \dfrac{ \sqrt{6} + \sqrt{7} }{ \sqrt{7} - \sqrt{6} } }$

$\purple{\bold{\boxed{\implies \: \dfrac{1}{x} = \dfrac{ \sqrt{7} - \sqrt{6} }{ \sqrt{6} + \sqrt{7} }}}}$

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Need to find:

${(x + \dfrac{1}{x} )}^{2} =?$

Substituting the values of x and $\dfrac{1}{x }$ in the given equation we get,

$= ( { \dfrac{ \sqrt{6} + \sqrt{7} }{ \sqrt{7} - \sqrt{6} } + \dfrac{ \sqrt{7} - \sqrt{6} }{ \sqrt{6} + \sqrt{7} } )}^{2}$

$= {( \dfrac{( \sqrt{6} + \sqrt{7} )( \sqrt{7 + \sqrt{6}) } }{ (\sqrt{7} - \sqrt{6} )( \sqrt{7} + \sqrt{6} )} + \dfrac{( \sqrt{7} - \sqrt{6})( \sqrt{7} - \sqrt{6}) }{( \sqrt{6} + \sqrt{7} )( \sqrt{7} - \sqrt{6} )}]}^{2}$

$= {( \dfrac{ {( \sqrt{7} + \sqrt{6} )}^{2} }{ { \sqrt{7} }^{2} - { \sqrt{6} }^{2} } + \dfrac{ {( \sqrt{7} - \sqrt{6} )}^{2} }{ { \sqrt{7} }^{2} - { \sqrt{6} }^{2} } )}^{2}$

$={( 7 + 6 + 2 \sqrt{42} + 7 + 6 - 2 \sqrt{42}) }^ {2}$

$= {(14 + 12)}^{2}$

$={ 26}^{2}$

$\red{\large{\bold{ = 676}}}$

Answer 2

$\large \sf{ \green{Solution : -}}$

$\sf{ \blue{Substituting \: the \: values \: of \: x \: and  \: \dfrac{1}{x }}} \\ \sf{ \blue{in \: the \: given \: equation \: we \: get,}}$

$= ( { \dfrac{ \sqrt{6} + \sqrt{7} }{ \sqrt{7} - \sqrt{6} } + \dfrac{ \sqrt{7} - \sqrt{6} }{ \sqrt{6} + \sqrt{7} } )}^{2} \\ \\ = {( \dfrac{( \sqrt{6} + \sqrt{7} )( \sqrt{7 + \sqrt{6}) } }{ (\sqrt{7} - \sqrt{6} )( \sqrt{7} + \sqrt{6} )} + \dfrac{( \sqrt{7} - \sqrt{6})( \sqrt{7} - \sqrt{6}) }{( \sqrt{6} + \sqrt{7} )( \sqrt{7} - \sqrt{6} )}]} \\ \\ = {( \dfrac{ {( \sqrt{7} + \sqrt{6} )}^{2} }{ { \sqrt{7} }^{2} - { \sqrt{6} }^{2} } + \dfrac{ {( \sqrt{7} - \sqrt{6} )}^{2} }{ { \sqrt{7} }^{2} - { \sqrt{6} }^{2} } )}^{2} \\ \\ ={( 7 + 6 + 2 \sqrt{42} + 7 + 6 - 2 \sqrt{42}) }^ {2} \\ \\ = {(14 + 12)}^{2} \\ \\ ={ 26}^{2} \\ \\ \red{\large{\bold{ \underline{ = 676}}}}$