The value of sqrt{12 - sqrt{68 + 48sqrt{2}}} | vedclass

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Question

(B) We are given the expression $\sqrt{12 - \sqrt{68 + 48\sqrt{2}}}$.First,simplify the inner radical $\sqrt{68 + 48\sqrt{2}}$.Note that $68 + 48\sqrt

Vedclass · Accepted Answer

$2 - \sqrt{2}$

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(B) We are given the expression $\sqrt{12 - \sqrt{68 + 48\sqrt{2}}}$.First,simplify the inner radical $\sqrt{68 + 48\sqrt{2}}$.Note that $68 + 48\sqrt{2} = 6^2 + (4\sqrt{2})^2 + 2 	imes 6 	imes 4\sqrt{2} = (6 + 4\sqrt{2})^2$.Thus,$\sqrt{68 + 48\sqrt{2}} = 6 + 4\sqrt{2}$.Substituting this back into the original expression,we get $\sqrt{12 - (6 + 4\sqrt{2})} = \sqrt{6 - 4\sqrt{2}}$.Now,express $6 - 4\sqrt{2}$ as a perfect square: $6 - 4\sqrt{2} = 2^2 + (\sqrt{2})^2 - 2 	imes 2 	imes \sqrt{2} = (2 - \sqrt{2})^2$.Therefore,$\sqrt{6 - 4\sqrt{2}} = \sqrt{(2 - \sqrt{2})^2} = 2 - \sqrt{2}$.

The value of $\sqrt{12 - \sqrt{68 + 48\sqrt{2}}} = $

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