Given: x2 – 2x + 1 = 0 and x = 1
On substituting x = 1 in L.H.S. of the given equation, we get:
(1)2 – 2(1) + 1
= 1 – 2 + 1
= 0
= R.H.S
Hence, L.H.S. = R.H.S.
x = 1 satisfies the given equation.
Therefore, x = 1 is a root of the given quadratic equation.