TS EAMCET · Maths · Basic of Mathematics
\(\sqrt{12-\sqrt{68+48 \sqrt{2}}}\) is equal to :
- A \(\sqrt{2}-3\)
- B \(2+\sqrt{2}\)
- C \(2-\sqrt{2}\)
- D \(6-2 \sqrt{8}\)
Answer & Solution
Correct Answer
(C) \(2-\sqrt{2}\)
Step-by-step Solution
Detailed explanation
\(\sqrt{12-\sqrt{68+48 \sqrt{2}}}\) \(=\sqrt{12-\sqrt{(6)^2+(4 \sqrt{2})^2+2 \times 6 \times 4 \sqrt{2}}}\) \(=\sqrt{12-\sqrt{(6+4 \sqrt{2})^2}}\) \(=\sqrt{12-6-4 \sqrt{2}}=\sqrt{6-4 \sqrt{2}}\) \(=\sqrt{(2)^2+(\sqrt{2})^2-2 \cdot \sqrt{2} \cdot 2}\)…
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