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

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Question

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

Vedclass · Accepted Answer

$2-\sqrt{2}$

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(C) Given expression: $\sqrt{12-\sqrt{68+48 \sqrt{2}}}$We simplify the inner radical $\sqrt{68+48 \sqrt{2}}$:$68+48 \sqrt{2} = 68+2 	imes 24 	imes \sqrt{2} = 68+2 	imes 6 	imes 4 \sqrt{2}$$= (6)^2 + (4 \sqrt{2})^2 + 2 	imes 6 	imes 4 \sqrt{2} = (6+4 \sqrt{2})^2$So,$\sqrt{68+48 \sqrt{2}} = 6+4 \sqrt{2}$Now substitute this back into the expression:$\sqrt{12-(6+4 \sqrt{2})} = \sqrt{6-4 \sqrt{2}}$$= \sqrt{6-2 	imes 2 	imes 2 \sqrt{2}} = \sqrt{6-2 	imes 2 	imes \sqrt{8}}$$= \sqrt{(2)^2 + (\sqrt{2})^2 - 2 	imes 2 	imes \sqrt{2}} = \sqrt{(2-\sqrt{2})^2}$$= 2-\sqrt{2}$

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

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