# Find a quadratic polynomial whose zeros are âˆš2+1 and 1Ã·âˆš2+1 .

$(\sqrt{2}+1) , \frac{1}{\sqrt{2}+1} are zeroes of the polynomial, so \\ (x-(\sqrt{2}+1)), (x - \frac{1}{\sqrt{2}+1}) are factors of the polynomial \\ (x-\sqrt{2}-1) (x - \frac{1}{\sqrt{2}+1})=0\\ (x-(\sqrt{2}+1)) (x(\sqrt{2}+1) -1) \\ (\sqrt{2}+1) x^2 -x - (\sqrt{2}+1)^2 x+ (\sqrt{2}+1) =0\\ (\sqrt{2}+1) x^2 -x -2x -x - 2\sqrt{2} x+ (\sqrt{2}+1) =0\\ (\sqrt{2}+1) x^2 -(4+ 2\sqrt{2}) x+ (\sqrt{2}+1) =0 \\ (\sqrt{2}+1) x^2 -2\sqrt{2}(\sqrt{2}+1) x+ (\sqrt{2}+1) =0 \\ x^2 -2\sqrt{2} x+ 1 =0 \\$

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