Solve the inequality −x² + 4x − 4 < 0.

Step-by-step solution

1. 1

   Factor out the leading minus sign to expose the structure.

   $-x^2+4x-4=-\left(x^2-4x+4\right)$

   The bracket is a familiar perfect square trinomial.

2. 2

   Recognise the perfect square.

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

3. 3

   Use the sign of a square. For every real $x$, $(x-2)^2\ge 0$, so $-(x-2)^2\le 0$. The expression is therefore never positive; the only question left is where it equals zero.

4. 4

   Locate the equality case. $-(x-2)^2=0$ exactly when $x=2$. Since the inequality is strict ($<0$, not $\le 0$), that single point must be excluded.

5. 5

   Write the solution set.

   $\boxed{x\in(-\infty,2)\cup(2,\infty),\ \text{i.e. all real }x\neq 2}$

6. 6

   Check the discriminant as a cross-check. For $-x^2+4x-4$ the discriminant is $4^2-4(-1)(-4)=16-16=0$, so the parabola touches the $x$-axis at one point and, opening downward ($a=-1<0$), lies strictly below it everywhere else — exactly the conclusion reached above. Spot values: at $x=0$ the value is $-4<0$, at $x=2$ it is $0$, at $x=5$ it is $-9<0$.