|2x – 3| < 1
Square both sides
⇒ (2x – 3)2 < 12
⇒ 4x2 – 12x + 9 < 1
⇒ 4x2 – 12x + 8 < 0
Divide throughout by 4
⇒ x2 – 3x + 2 < 0
⇒ x2 – 2x – x + 2 < 0
⇒ x(x – 2) – 1(x – 2) < 0
⇒ (x – 1)(x – 2) < 0
Observe that when x is greater than 2 (x – 1)(x – 2) 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 2 for (x – 1)(x – 2) to be negative
Hence x ∈ (1, 2)
Hence the solution set of |2x – 3| < 1 is (1, 2)
