If x = sqrt 7 + sqrt 3 and xy = 4,then {x^4} | vedclass

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Question

(b) $x = \sqrt 7 + \sqrt 3 ,\,\,xy = 4$

$ \Rightarrow $$y = {4 \over x} = {4 \over {\sqrt 7 + \sqrt 3 }} = {{4(\sqrt 7 - \sqrt 3 )} \over {7 - 3}}

Vedclass · Accepted Answer

$368$

Vedclass · Answer

(b) $x = \sqrt 7 + \sqrt 3 ,\,\,xy = 4$

$ \Rightarrow $$y = {4 \over x} = {4 \over {\sqrt 7 + \sqrt 3 }} = {{4(\sqrt 7 - \sqrt 3 )} \over {7 - 3}} = \sqrt 7 - \sqrt 3 $

${x^4} + {y^4} = {({x^2} + {y^2})^2} - 2{x^2}{y^2}$

$ = {[{(x + y)^2} - 2xy]^2} - 2{(xy)^2} = {[{(2\sqrt 7 )^2} - 8]^2} - 2\,.\,{4^2} = 368$.

If $x = \sqrt 7 + \sqrt 3 $ and $xy = 4,$then ${x^4} + {y^4}=$

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