{{12} over {3 + sqrt 5 - 2sqrt 2 }} = | vedclass

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Question

(c) ${{12} \over {3 + \sqrt 5 - 2\sqrt 2 }} = {{12\,[(3 - 2\sqrt 2 ) - \sqrt 5 ]} \over {[(3 - 2\sqrt 2 ) + \sqrt 5 ][(3 - 2\sqrt 2 ) - \sqrt 5 ]}}$

Vedclass · Accepted Answer

$1 + \sqrt 5 + \sqrt {10} - \sqrt 2 $

Vedclass · Answer

(c) ${{12} \over {3 + \sqrt 5 - 2\sqrt 2 }} = {{12\,[(3 - 2\sqrt 2 ) - \sqrt 5 ]} \over {[(3 - 2\sqrt 2 ) + \sqrt 5 ][(3 - 2\sqrt 2 ) - \sqrt 5 ]}}$

$ = {{12\,(3 - 2\sqrt 2 - \sqrt 5 )} \over {{{(3 - 2\sqrt 2 )}^2} - 5}} = {{12\,(3 - 2\sqrt 2 - \sqrt 5 )} \over {17 - 12\sqrt 2 - 5}}$

$ = {{(3 - 2\sqrt 2 - \sqrt 5 )} \over {1 - \sqrt 2 }} = {{(\sqrt 5 + 2\sqrt 2 - 3)\,(\sqrt 2 + 1)} \over {(\sqrt 2 - 1)\,(\sqrt 2 + 1)}}$

$ = {{\sqrt {10} + 4 - 3\sqrt 2 + \sqrt 5 + 2\sqrt 2 - 3} \over {2 - 1}} = 1 + \sqrt 5 + \sqrt {10} - \sqrt 2 $.

${{12} \over {3 + \sqrt 5 - 2\sqrt 2 }} = $

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