Calculus
posted by Zukii .
solve
x2x+1>3

x^2  x + 1 > 3
or
x^2 + x  1 > 3
case 1:
x^2 x  2 > 0
(x2)(x+1) > 0
critical values are 1 and 2
(if we look at the parabola y = x^2  x  2, the xintercepts are 1 and 2 and we want all values above the xaxis)
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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