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$\sqrt{12-\sqrt{68+48 \sqrt{2}}}$ is equal to :
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Verified Answer
The correct answer is:
$2-\sqrt{2}$
$\sqrt{12-\sqrt{68+48 \sqrt{2}}}$
$=\sqrt{12-\sqrt{(6)^2+(4 \sqrt{2})^2+2 \times 6 \times 4 \sqrt{2}}}$
$=\sqrt{12-\sqrt{(6+4 \sqrt{2})^2}}$
$=\sqrt{12-6-4 \sqrt{2}}=\sqrt{6-4 \sqrt{2}}$
$=\sqrt{(2)^2+(\sqrt{2})^2-2 \cdot \sqrt{2} \cdot 2}$
$=\sqrt{(2-\sqrt{2})^2}=2-\sqrt{2}$
$=\sqrt{12-\sqrt{(6)^2+(4 \sqrt{2})^2+2 \times 6 \times 4 \sqrt{2}}}$
$=\sqrt{12-\sqrt{(6+4 \sqrt{2})^2}}$
$=\sqrt{12-6-4 \sqrt{2}}=\sqrt{6-4 \sqrt{2}}$
$=\sqrt{(2)^2+(\sqrt{2})^2-2 \cdot \sqrt{2} \cdot 2}$
$=\sqrt{(2-\sqrt{2})^2}=2-\sqrt{2}$
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