sqrt{12-sqrt{68+48 sqrt{2}}} is equal to : | vedclass

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Question

(C) Given expression: $\sqrt{12-\sqrt{68+48 \sqrt{2}}}$First,simplify the inner radical: $\sqrt{68+48 \sqrt{2}} = \sqrt{68+2 	imes 24 \sqrt{2}} = \sq

Vedclass · Accepted Answer

$2-\sqrt{2}$

Vedclass · Answer

(C) Given expression: $\sqrt{12-\sqrt{68+48 \sqrt{2}}}$First,simplify the inner radical: $\sqrt{68+48 \sqrt{2}} = \sqrt{68+2 	imes 24 \sqrt{2}} = \sqrt{68+2 	imes 6 	imes 4 \sqrt{2}}$Since $(6)^2 + (4 \sqrt{2})^2 = 36 + 32 = 68$,we have $\sqrt{68+48 \sqrt{2}} = \sqrt{(6+4 \sqrt{2})^2} = 6+4 \sqrt{2}$Substitute this back: $\sqrt{12-(6+4 \sqrt{2})} = \sqrt{6-4 \sqrt{2}}$Now,express $6-4 \sqrt{2}$ as a perfect square: $6-4 \sqrt{2} = 4 + 2 - 2 	imes 2 	imes \sqrt{2} = (2)^2 + (\sqrt{2})^2 - 2 	imes 2 	imes \sqrt{2} = (2-\sqrt{2})^2$Therefore,$\sqrt{6-4 \sqrt{2}} = \sqrt{(2-\sqrt{2})^2} = 2-\sqrt{2}$

$\sqrt{12-\sqrt{68+48 \sqrt{2}}}$ is equal to :

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