What is |x^2-2x-3|>1 ?

The solution to |x^2-2x-3|>1 is x sqrt(5)+1

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|x^2-2x-3|>1

x^{2} - 2 x - 3 > 1

x < - 5 + 1 or - 3 + 1 < x < 3 + 1 or x > 5 + 1

Solution steps

x^{2} - 2 x - 3 > 1

Apply absolute rule: If

∣ u ∣ > a, a > 0

then

u < - a

u > a

x^{2} - 2 x - 3 < - 1 or x^{2} - 2 x - 3 > 1

x^{2} - 2 x - 3 < - 1 or x^{2} - 2 x - 3 > 1

- 3 + 1 < x < 3 + 1 or (x < - 5 + 1 or x > 5 + 1)

Merge Overlapping Intervals

x < - 5 + 1 or - 3 + 1 < x < 3 + 1 or x > 5 + 1

∣ x - 2 ∣ + ∣ x - 3 ∣ + ∣ x - 4 ∣ > 6

∣ 5 x - 2 ∣ < 3 x + 4

\frac{2 x + 2}{6} > 4

∣ x - 5 ∣ + 2 x - 1 \geq ∣ x + 4 ∣

x^{2} - 4 x + 3 \geq x^{2} - 6 x + 8

What is |x^2-2x-3|>1 ?
The solution to |x^2-2x-3|>1 is x<-sqrt(5)+1\lor-sqrt(3)+1<x<sqrt(3)+1\lor x>sqrt(5)+1