Substituting \(x = 1\) into the equation \(x^2 + x + 1 = 0\): \[ (1)^2 + 1 + 1 = 1 + 1 + 1 = 3 \quad [ā 0; \;\therefore\; the equation is not satisfied by 1] \] Substituting \(x = -1\) into the equation \(x^2 + x + 1 = 0\): \[ (-1)^2 + (-1) + 1 = 1 - 1 + 1 = 1 \quad [ā 0; \;\therefore\; the equation is not satisfied by -1] \] Therefore, 1 and -1 cannot be the roots of the quadratic equation \(x^2 + x + 1 = 0\).