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

The solution to |z^2-1|<= 2 is -sqrt(3)<= z<= sqrt(3)

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|z^2-1|<= 2

z^{2} - 1 \leq 2

- 3 \leq z \leq 3

Solution steps

z^{2} - 1 \leq 2

Apply absolute rule: If

∣ u ∣ \leq a, a > 0

then

- a \leq u \leq a

- 2 \leq z^{2} - 1 \leq 2

z^{2} - 1 \geq - 2 and z^{2} - 1 \leq 2

True for all z and - 3 \leq z \leq 3

Merge Overlapping Intervals

- 3 \leq z \leq 3

\frac{x ^{2}}{x - 1} > 0

- y^{2} x^{4} + \frac{4}{3} x^{3} - 2 x^{2} + 3 x - 1

\frac{2 x - 6}{x + 1} \leq 1

25 x^{2} - 100

24 - 0.25 (x - (- 9))^{2} > 11.75