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

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Question

(B) First,simplify the inner nested radical $\sqrt{68 + 48\sqrt{2}}$.We can write $48\sqrt{2} = 2 	imes 24 	imes \sqrt{2} = 2 	imes \sqrt{576 	ime

Vedclass · Accepted Answer

$2 - \sqrt{2}$

Vedclass · Answer

(B) First,simplify the inner nested radical $\sqrt{68 + 48\sqrt{2}}$.We can write $48\sqrt{2} = 2 	imes 24 	imes \sqrt{2} = 2 	imes \sqrt{576 	imes 2} = 2 	imes \sqrt{1152}$.Alternatively,express $68 + 48\sqrt{2}$ in the form $(a + b\sqrt{2})^2 = a^2 + 2b^2 + 2ab\sqrt{2}$.Comparing,$2ab = 48 \implies ab = 24$. Let $a = 6, b = 4$. Then $a^2 + 2b^2 = 36 + 2(16) = 36 + 32 = 68$.So,$\sqrt{68 + 48\sqrt{2}} = \sqrt{(6 + 4\sqrt{2})^2} = 6 + 4\sqrt{2}$.Now,substitute this into the original expression: $\sqrt{12 - (6 + 4\sqrt{2})} = \sqrt{6 - 4\sqrt{2}}$.We need to simplify $\sqrt{6 - 4\sqrt{2}}$. Let $\sqrt{6 - 4\sqrt{2}} = \sqrt{x} - \sqrt{y}$.Squaring both sides: $6 - 4\sqrt{2} = x + y - 2\sqrt{xy}$.Thus,$x + y = 6$ and $2\sqrt{xy} = 4\sqrt{2} \implies \sqrt{xy} = 2\sqrt{2} = \sqrt{8} \implies xy = 8$.The numbers $x$ and $y$ are $4$ and $2$.Therefore,$\sqrt{6 - 4\sqrt{2}} = \sqrt{4} - \sqrt{2} = 2 - \sqrt{2}$.

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

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