MCQ
If $x = 3 - \sqrt {5,} $ then ${{\sqrt x } \over {\sqrt 2 + \sqrt {(3x - 2)} }} = $
- A$5$
- B$\sqrt 5 $
- C$1/5$
- ✓$1/\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 $;
$\therefore {{\sqrt x } \over {\sqrt 2 + \sqrt {3x - 2} }} = {1 \over {\sqrt 5 }}$.
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