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

Step-by-step solution

1. 1

   Factor out $-1$ and recognise the perfect square.

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

   The bracket is a perfect square because $x^2-4x+4=x^2-2\cdot x\cdot 2+2^2$.

2. 2

   Reduce the inequality. The statement becomes

   $-(x-2)^2<0\iff(x-2)^2>0$

   Multiplying by $-1$ reversed the direction.

3. 3

   Solve the squared inequality. A square is always $\ge 0$ and equals $0$ only when its base does:

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

4. 4

   Write the solution set. Every real number except the repeated root:

   $x\in(-\infty,2)\cup(2,\infty)$

5. 5

   Check the excluded point and its neighbours. At $x=2$: $-4+8-4=0$, which is not $<0$, so $2$ is genuinely excluded $\checkmark$. At $x=1.9$: $-(0.01)=-0.01<0$ $\checkmark$. At $x=100$: $-(98)^2<0$ $\checkmark$. Had the inequality been $\le 0$ instead, the answer would have been all real numbers.