What is |x^2-1|<= |x+3| ?

The solution to |x^2-1|<= |x+3| is (-sqrt(17)+1)/2 <= x<= (sqrt(17)+1)/2

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|x^2-1|<= |x+3|

x^{2} - 1 \leq ∣ x + 3 ∣

\frac{- 17 + 1}{2} \leq x \leq \frac{17 + 1}{2}

Solution steps

x^{2} - 1 \leq ∣ x + 3 ∣

Find positive and negative intervals

x < - 3, - 3 \leq x \leq - 1, - 1 < x < 1, x \geq 1

Solve the inequality for each interval

No Solution or \frac{- 17 + 1}{2} \leq x \leq - 1 or - 1 < x < 1 or 1 \leq x \leq \frac{17 + 1}{2}

Merge Overlapping Intervals

\frac{- 17 + 1}{2} \leq x \leq \frac{17 + 1}{2}

x^{2} - 3 x - 4 < x + 1

∣ - 2 x + 2 ∣ - 1 > 5

x^{2} - 5 x \geq 6

∣ x + 2 ∣ > 3 x

x^{3} - 7 x + 6 \geq 19 x - x^{3} - 18

What is |x^2-1|<= |x+3| ?
The solution to |x^2-1|<= |x+3| is (-sqrt(17)+1)/2 <= x<= (sqrt(17)+1)/2