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STD 11 - basic of algoritham — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsSTD 11 - basic of algoritham1 Mark
MCQ
$\root 4 \of {(17 + 12\sqrt 2 )} = $
  • ✓
    $\sqrt 2 + 1$
  • B
    ${2^{1/4}}(\sqrt 2 + 1)$
  • C
    $2\sqrt 2 + 1$
  • D
    None of these
✓

Answer

Correct option: A.
$\sqrt 2 + 1$
a
(a) $\sqrt {(17 + 12\sqrt 2 )} $$ = \sqrt {[{3^2} + {{(2\sqrt 2 )}^2} + 2.3.2\sqrt 2 ]} = 3 + 2\sqrt 2 $

$\therefore \sqrt[4]{{(17 + 12\sqrt 2 )}} = \sqrt {(3 + 2\sqrt 2 )}  = \sqrt 2  + 1$

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