Solve the inequality 4x2 - 4x + 1 < 0.

Step-by-step solution

1. 1

   Look for a perfect square before factoring blindly. The outer terms are squares, $4x^2=(2x)^2$ and $1=1^2$, and the middle term is $-2\cdot(2x)\cdot 1 = -4x$. That is exactly the pattern $a^2-2ab+b^2$, so $4x^2-4x+1=(2x-1)^2.$

2. 2

   Check the discriminant to be sure. $\Delta = (-4)^2-4(4)(1) = 16-16 = 0$. A zero discriminant means a repeated root, which is another way of saying the quadratic is a perfect square with a single root $x=\tfrac12$.

3. 3

   Use the sign property of squares. For every real $x$, $(2x-1)^2 \ge 0$. A real square can be zero, but it can never be strictly negative, so the requirement $(2x-1)^2 < 0$ can never be met.

4. 4

   Conclude the solution set is empty. $4x^2-4x+1<0 \iff (2x-1)^2<0 \iff \varnothing.$ Geometrically the parabola opens upward and touches the $x$-axis at $x=\tfrac12$ without ever dipping below it.

5. 5

   Contrast the neighbouring cases so the strictness is not lost. $(2x-1)^2 \le 0$ has the single solution $x=\tfrac12$; $(2x-1)^2 > 0$ has everything except $x=\tfrac12$; and $(2x-1)^2 \ge 0$ is true for all real $x$. Only the strict $<$ gives the empty set.