x^₂ + x $-$ 1 = 0, x = 2

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Solution

Given: x² + x – 1 = 0 and x = 2
On substituting x = 2 in L.H.S. of the given equation, we get:
= (2)² + 2 - 1
= 4 + 2 - 1
=5
=> L.H.S. $\neq$ R.H.S.
x = 2 does not satisfy the given equation.
Therefore, x = 2 is not a root of the given quadratic equation.

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