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 $10 - \sqrt {24} - \sqrt {40} + \sqrt {60} = {(\sqrt a - \sqrt b + \sqrt c )^2}$
(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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