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Examveda

$$\sqrt {6 - 4\sqrt 3 + \sqrt {16 - 8\sqrt 3 } } $$       is equal to = ?

A. $${\text{1}} - \sqrt 3 $$

B. $$\sqrt 3 - 1$$

C. $${\text{2}}\left( {2 - \sqrt 3 } \right)$$

D. $${\text{2}}\left( {2 + \sqrt 3 } \right)$$

Answer: Option B

Solution(By Examveda Team)

$$\eqalign{ & \sqrt {6 - 4\sqrt 3 + \sqrt {16 - 8\sqrt 3 } } \cr & = \sqrt {6 - 4\sqrt 3 + \sqrt {12 + 4 - 8\sqrt 3 } } \cr & = \sqrt {6 - 4\sqrt 3 + \sqrt {{{\left( {2\sqrt 3 } \right)}^2} + {{\left( 2 \right)}^2} - 2 \times 2\sqrt 3 \times 2} } \cr & = \sqrt {6 - 4\sqrt 3 + \sqrt {{{\left( {2\sqrt 3 - 2} \right)}^2}} } \cr & = \sqrt {6 - 4\sqrt 3 + 2\sqrt 3 - 2} \cr & = \sqrt {{{\left( {\sqrt 3 } \right)}^2} + {{\left( 1 \right)}^2} - 2 \times \sqrt 3 \times 1} \cr & = \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} \cr & = \sqrt 3 - 1 \cr} $$

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