What is |x^2-x-2|>2 ?

The solution to |x^2-x-2|>2 is x (sqrt(17)+1)/2

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|x^2-x-2|>2

x^{2} - x - 2 > 2

x < \frac{- 17 + 1}{2} or 0 < x < 1 or x > \frac{17 + 1}{2}

Solution steps

x^{2} - x - 2 > 2

Apply absolute rule: If

∣ u ∣ > a, a > 0

then

u < - a

u > a

x^{2} - x - 2 < - 2 or x^{2} - x - 2 > 2

x^{2} - x - 2 < - 2 or x^{2} - x - 2 > 2

0 < x < 1 or (x < \frac{- 17 + 1}{2} or x > \frac{17 + 1}{2})

Merge Overlapping Intervals

x < \frac{- 17 + 1}{2} or 0 < x < 1 or x > \frac{17 + 1}{2}

∣ 5 - x ∣ + ∣ x - 13 ∣ < ∣ 2 - x ∣

∣ x ∣ + ∣ x - 4 ∣ > 4

(b) ∣ x + 2 ∣ + 2 \geq 0

\frac{( 3 x + 6 )}{9} \geq 5

4 ∣ x - 9 ∣ > 16

What is |x^2-x-2|>2 ?
The solution to |x^2-x-2|>2 is x<(-sqrt(17)+1)/2 \lor 0<x<1\lor x>(sqrt(17)+1)/2