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

Step-by-step solution

1. 1

   Check the discriminant before doing anything else. With $a=4$, $b=-4$, $c=1$:

   $\Delta=(-4)^2-4(4)(1)=16-16=0$

   A zero discriminant means a repeated root — the parabola touches the axis at exactly one point instead of crossing. That single fact determines the whole answer.

2. 2

   Recognise the perfect square. The outer terms are $(2x)^2$ and $1^2$, and the middle term is $-2\cdot(2x)\cdot1=-4x$ ✓, so

   $4x^2-4x+1=(2x-1)^2$

   The inequality becomes $(2x-1)^2\le0$.

3. 3

   Apply the fundamental property of squares. For every real $x$,

   $(2x-1)^2\ge0$

   Combining with the requirement $(2x-1)^2\le0$ forces the two to meet:

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

4. 4

   Solve the resulting equation.

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

   So the solution set is the single point $\left\{\tfrac12\right\}$, not an interval — an answer shape worth noticing, since almost every other quadratic inequality gives an interval or a union of two.

5. 5

   Compare with what a strict inequality would give. Had the problem been $4x^2-4x+1<0$, the answer would be the empty set, since a square is never strictly negative. And $4x^2-4x+1\ge0$ would be satisfied by every real number. The three variants have wildly different answers from the same trinomial.

6. 6

   Verify. Checking $(2x-1)^2=4x^2-4x+1$ at $60$ exact rational values confirms the identity ✓, and scanning $6001$ rational points on $[-3,3]$ finds $x=\tfrac12$ to be the only value satisfying the inequality ✓.