MCQ
$\sqrt {[10 - \sqrt {(24)} - \sqrt {(40)} + \sqrt {(60)} ]} = $
  • A
    $\sqrt 5 + \sqrt 3 + \sqrt 2 $
  • $\sqrt 5 + \sqrt 3 - \sqrt 2 $
  • C
    $\sqrt 5 - \sqrt 3 + \sqrt 2 $
  • D
    $\sqrt 2 + \sqrt 3 - \sqrt 5 $

Answer

Correct option: B.
$\sqrt 5 + \sqrt 3 - \sqrt 2 $
b
(b) Let $10 - \sqrt {24} - \sqrt {40} + \sqrt {60} = {(\sqrt a - \sqrt b + \sqrt c )^2}$

$ = a + b + c - 2\sqrt {ab} - 2\sqrt {bc} + 2\sqrt {ca} $

$a,b,c > 0$. Then $a + b + c = 10,$

$ab = 6$, $bc = 10,$$ca = 15$

${a^2}{b^2}{c^2} = 900$==> $abc = 30$ $( \ne \pm 30)$.

So, $a = 3,\,\,\,b = 2,\,\,c = 5$

Therefore, $\sqrt {(10 - \sqrt {24} - \sqrt {40} + \sqrt {60} )} = \pm (\sqrt 3 + \sqrt 5 - \sqrt 2 )$

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