|x – 1| < 2
Square both sides
⇒ (x – 1)2 < 4
⇒ x 2 – 2x + 1 < 4
⇒ x 2 – 2x – 3 < 0
⇒ x 2 – 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)
