|x – 1| < 2

Square both sides

⇒ (x – 1)² < 4

⇒ x² – 2x + 1 < 4

⇒ x² – 2x – 3 < 0

⇒ x² – 3x + x – 3 < 0

⇒ x(x – 3) + 1(x – 3) < 0

⇒ (x + 1)(x – 3) < 0

Observe that when x > 3 (x – 3)(x + 1) is positive

And for each root the sign changes hence

We want less than 0 that is negative part

Hence x should be between -1 and 3 for (x – 3)(x + 1) to be negative

Hence x ∈ (-1, 3)

Hence solution set for |x – 1| < 2 is (-1, 3)