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

The solution to |x^2-1|<= |2x-1| is -sqrt(3)-1<= x<= 0\lor sqrt(3)-1<= x<= 2

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

x^{2} - 1 \leq ∣ 2 x - 1 ∣

- 3 - 1 \leq x \leq 0 or 3 - 1 \leq x \leq 2

Interval Notation

[- 3 - 1, 0] \cup [3 - 1, 2]

Decimal Notation

- 2.73205 \dots \leq x \leq 0 or 0.73205 \dots \leq x \leq 2

Solution steps

x^{2} - 1 \leq ∣ 2 x - 1 ∣

Find positive and negative intervals

x \leq - 1, - 1 < x < \frac{1}{2}, \frac{1}{2} \leq x < 1, x \geq 1

Solve the inequality for each interval

- 3 - 1 \leq x \leq - 1 or - 1 < x \leq 0 or 3 - 1 \leq x < 1 or 1 \leq x \leq 2

Merge Overlapping Intervals

- 3 - 1 \leq x \leq 0 or 3 - 1 \leq x \leq 2

\frac{x}{( x ^{2} - x - 1 )} < 2

∣ 4 x - 3 ∣ + 2 < 10

\frac{2}{∣ x - 4 ∣} > 1

lo g_{2} (∣ x - 2 ∣) > lo g_{2} (∣ x + 4 ∣)

∣ x + 1 ∣^{2} + 2 ∣ x + 2 ∣ \geq 2

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