Solve (2x − 3)² − (x − 1)² = 0.

Step-by-step solution

1. 1

   Recognise the difference-of-squares structure. Both terms are already perfect squares of linear expressions, so

   $A^2 - B^2 = (A - B)(A + B) \qquad \text{with } A = 2x - 3, \ B = x - 1$

   No expansion is needed; the equation factors immediately.

2. 2

   Form the first factor, distributing the minus sign.

   $A - B = (2x - 3) - (x - 1) = 2x - 3 - x + 1 = x - 2$

   The subtraction of $-1$ becomes $+1$; dropping that sign flip is the usual error and would give $x - 4$ instead.

3. 3

   Form the second factor.

   $A + B = (2x - 3) + (x - 1) = 3x - 4$

   So the equation becomes

   $(x - 2)(3x - 4) = 0$

4. 4

   Solve each linear factor.

   $x - 2 = 0 \ \Rightarrow \ x = 2, \qquad 3x - 4 = 0 \ \Rightarrow \ x = \frac{4}{3}$

5. 5

   Verify both. At $x = 2$: $(4-3)^2 - (2-1)^2 = 1 - 1 = 0 \ \checkmark$. At $x = \tfrac43$: $\left(\tfrac83 - 3\right)^2 - \left(\tfrac43 - 1\right)^2 = \tfrac19 - \tfrac19 = 0 \ \checkmark$. Geometrically, $(2x-3)^2 = (x-1)^2$ means $|2x-3| = |x-1|$, and the two roots are exactly the cases $2x - 3 = x - 1$ and $2x - 3 = -(x-1)$.