x^2 - x + 1 > 3
or
-x^2 + x - 1 > 3
case 1:
x^2 -x - 2 > 0
(x-2)(x+1) > 0
critical values are -1 and 2
(if we look at the parabola y = x^2 - x - 2, the x-intercepts are -1 and 2 and we want all values above the x-axis)
so x < -1 OR x > 2
case 2:
x^2 -x + 4 < 0
the corresponding equation x^2 - x + 4 = 0 has no real roots, so no value of x satisfies our inequation.
so x < -1 OR x > 2
solve
|x2-x+1|>3
1 answer