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March 26, 2017

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solve
|x2-x+1|>3

  • Calculus - ,

    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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