Use factoring to solve (2x - 1)(2x - 1) - x * x = 0.

Step-by-step solution

1. 1

   Rewrite the repeated products as squares. The two identical factors and the two identical $x$'s are just squares written out longhand:

   $(2x-1)(2x-1) - x \cdot x = (2x-1)^2 - x^2 = 0$

   Seeing the squares is the whole trick here; expanding into $3x^2 - 4x + 1$ also works, but it throws away the structure.

2. 2

   Match the difference-of-squares pattern. The identity is

   $a^2 - b^2 = (a - b)(a + b)$

   with $a = 2x - 1$ and $b = x$. Note that $a$ is itself a binomial, which is fine: the pattern only cares that each term is a perfect square.

3. 3

   Apply the pattern and simplify each bracket.

   $(2x-1)^2 - x^2 = \big[(2x-1) - x\big]\big[(2x-1) + x\big]$

   $(2x-1) - x = x - 1, \qquad (2x-1) + x = 3x - 1$

   $\Longrightarrow \quad (x - 1)(3x - 1) = 0$

   Expanding back gives $3x^2 - 4x + 1$, which agrees with expanding the original expression.

4. 4

   Use the zero-product property. A product is zero exactly when one factor is zero:

   $x - 1 = 0 \quad \Longrightarrow \quad x = 1$

   $3x - 1 = 0 \quad \Longrightarrow \quad x = \frac{1}{3}$

5. 5

   Check both roots in the original form. For $x = 1$: $(2-1)^2 - 1^2 = 1 - 1 = 0$. For $x = \tfrac13$: $\left(\tfrac23 - 1\right)^2 - \left(\tfrac13\right)^2 = \tfrac19 - \tfrac19 = 0$. Both work, so the solution set is $\left\{1, \tfrac13\right\}$.