If x = 3 - sqrt {5,} then {{sqrt x } over {s | vedclass

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Question

(d) $x = 3 - \sqrt 5 $

$\sqrt x = \sqrt {3 - \sqrt 5 } = {1 \over {\sqrt 2 }}\,.\sqrt {6 - 2\sqrt 5 } = {1 \over {\sqrt 2 }}(\sqrt 5 - 1)$  &n

Vedclass · Accepted Answer

$1/\sqrt 5 $

Vedclass · Answer

(d) $x = 3 - \sqrt 5 $

$\sqrt x = \sqrt {3 - \sqrt 5 } = {1 \over {\sqrt 2 }}\,.\sqrt {6 - 2\sqrt 5 } = {1 \over {\sqrt 2 }}(\sqrt 5 - 1)$         $3x - 2 = 9 - 3\sqrt 5 - 2 = 7 - 3\sqrt 5 = {{14 - 6\sqrt 5 } \over 2}$

= ${{{{(3 - \sqrt 5 )}^2}} \over 2}$;

$ \Rightarrow $ $\sqrt 2 + \sqrt {3x - 2} = \sqrt 5 \,.\,\sqrt x $;

$	herefore {{\sqrt x } \over {\sqrt 2 + \sqrt {3x - 2} }} = {1 \over {\sqrt 5 }}$.

If $x = 3 - \sqrt {5,} $ then ${{\sqrt x } \over {\sqrt 2 + \sqrt {(3x - 2)} }} = $

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