(A) We have $\sqrt{17 + 12\sqrt{2}} = \sqrt{3^2 + (2\sqrt{2})^2 + 2 \times 3 \times 2\sqrt{2}}$.This simplifies to $\sqrt{(3 + 2\sqrt{2})^2} = 3 + 2\s

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(A) We have $\sqrt{17 + 12\sqrt{2}} = \sqrt{3^2 + (2\sqrt{2})^2 + 2 \times 3 \times 2\sqrt{2}}$.This simplifies to $\sqrt{(3 + 2\sqrt{2})^2} = 3 + 2\sqrt{2}$.Therefore,$\sqrt[4]{17 + 12\sqrt{2}} = \sqrt{3 + 2\sqrt{2}}$.Since $3 + 2\sqrt{2} = 2 + 1 + 2\sqrt{2} = (\sqrt{2})^2 + 1^2 + 2\sqrt{2} = (\sqrt{2} + 1)^2$,we get $\sqrt{3 + 2\sqrt{2}} = \sqrt{(\sqrt{2} + 1)^2} = \sqrt{2} + 1$.