Find the zeroes of zeros x^2 – (√2+1)x+√2

The given equation is x^2 – (√2+1)x+√2. On applying trial and error method for x = 1, we get,

1^2 – (√2+1)1+√2 = 1 – √2 + 1 + √2 = 0. Thus, x = 1 is the zero for x^2 – (√2+1)x+√2.

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