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

The solution to x^2-1<= |x+1| is -1<= x<= 2

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

x^{2} - 1 \leq ∣ x + 1 ∣

- 1 \leq x \leq 2

Interval Notation

[- 1, 2]

Solution steps

x^{2} - 1 \leq ∣ x + 1 ∣

Find positive and negative intervals

x < - 1, x \geq - 1

Solve the inequality for each interval

No Solution or - 1 \leq x \leq 2

Merge Overlapping Intervals

- 1 \leq x \leq 2

0.9 t = 1 + t

0 \leq 6 - \frac{3}{2} x \leq 9

56 x^{2} - 56 x + 14 = 0

2 x^{2} + 12 x + 18