What is |(x+1)^2-3|>0 ?

The solution to |(x+1)^2-3|>0 is x sqrt(3)-1

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|(x+1)^2-3|>0

(x + 1)^{2} - 3 > 0

x < - 3 - 1 or - 3 - 1 < x < 3 - 1 or x > 3 - 1

Interval Notation

(- \infty, - 3 - 1) \cup (- 3 - 1, 3 - 1) \cup (3 - 1, \infty)

Decimal Notation

x < - 2.73205 \dots or - 2.73205 \dots < x < 0.73205 \dots or x > 0.73205 \dots

Solution steps

(x + 1)^{2} - 3 > 0

Apply absolute rule: If

∣ u ∣ > 0

then

u < 0

u > 0

(x + 1)^{2} - 3 < 0 or (x + 1)^{2} - 3 > 0

(x + 1)^{2} - 3 < 0 or (x + 1)^{2} - 3 > 0

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

Merge Overlapping Intervals

x < - 3 - 1 or - 3 - 1 < x < 3 - 1 or x > 3 - 1

\frac{x - 3}{4} < \frac{10}{3}

3 t^{2} + t - 7 = 0

3 x + 5 \leq - 7 x + 8

2 - \frac{6}{5}

ln (e^{- 1})

What is |(x+1)^2-3|>0 ?
The solution to |(x+1)^2-3|>0 is x<-sqrt(3)-1\lor-sqrt(3)-1<x<sqrt(3)-1\lor x>sqrt(3)-1