What is |x^2+x|+x-1<0 ?

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

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|x^2+x|+x-1<0

x^{2} + x + x - 1 < 0

- 2 - 1 < x < 2 - 1

Interval Notation

(- 2 - 1, 2 - 1)

Decimal Notation

- 2.41421 \dots < x < 0.41421 \dots

Solution steps

x^{2} + x + x - 1 < 0

Find positive and negative intervals

x \leq - 1, - 1 < x < 0, x \geq 0

Solve the inequality for each interval

- 2 - 1 < x \leq - 1 or - 1 < x < 0 or 0 \leq x < 2 - 1

Merge Overlapping Intervals

- 2 - 1 < x < 2 - 1

∣ u - 6 ∣ < 5

\frac{1}{x} \geq ∣ x + 2 ∣

\frac{( x ^{2} - 1 )}{( x ^{2} + x + 1 )} > 1

∣ x + 3 ∣ < 0.01

∣ x - 1 ∣ > 2 ∣ x - 4 ∣