Solve the inequality 4x^2 - 4x + 1 > 0.

Step-by-step solution

1. 1

   Match the perfect-square pattern with a non-unit leading term. Here $4x^2=(2x)^2$ and $1=1^2$, and the middle term checks out:

   $2\cdot(2x)\cdot 1=4x\quad\Longrightarrow\quad 4x^2-4x+1=(2x-1)^2$

2. 2

   Confirm with the discriminant. $\Delta=(-4)^2-4(4)(1)=16-16=0$, so there is exactly one repeated root and no sign change.

3. 3

   Reduce to a statement about a square.

   $(2x-1)^2>0\iff 2x-1\neq 0$

4. 4

   Solve the excluded equation.

   $2x-1=0\quad\Longrightarrow\quad x=\frac12$

   so every real number other than $\tfrac12$ satisfies the inequality.

5. 5

   Check and compare with the unit-coefficient version. At $x=\tfrac12$: $1-2+1=0$, not $>0$ $\checkmark$. At $x=0$: $1>0$ $\checkmark$. The structure is identical to $x^2-2x+1>0$, but the repeated root moves from $1$ to $\tfrac12$ because the base is $2x-1$ rather than $x-1$.