What is |x^2-1|<= 4 ?

The solution to |x^2-1|<= 4 is -sqrt(5)<= x<= sqrt(5)

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|x^2-1|<= 4

x^{2} - 1 \leq 4

- 5 \leq x \leq 5

Solution steps

x^{2} - 1 \leq 4

Apply absolute rule: If

∣ u ∣ \leq a, a > 0

then

- a \leq u \leq a

- 4 \leq x^{2} - 1 \leq 4

x^{2} - 1 \geq - 4 and x^{2} - 1 \leq 4

True for all x and - 5 \leq x \leq 5

Merge Overlapping Intervals

- 5 \leq x \leq 5

∣ x - 2 ∣ + ∣ x - 4 ∣ > 6

x^{2} - ∣ 3 x + 2 ∣ + x \geq 0

\frac{x + 5}{∣ 1 - x ∣} > 1

∣ 2 x - 10 ∣ > 10

∣ x ∣^{x^{2} - x - 2} < 1