Posted by **Zukii** on Thursday, January 13, 2011 at 11:13pm.

solve

|x2-x+1|>3

- Calculus -
**Reiny**, Friday, January 14, 2011 at 12:35am
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

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