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

Step-by-step solution

1. 1

   Spot the perfect square before reaching for the formula. The pattern $a^2-2ab+b^2$ with $a=x$, $b=1$ gives

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

   The discriminant confirms it: $\Delta=4-4=0$, the signature of a repeated root.

2. 2

   Rewrite the inequality.

   $(x-1)^2>0$

3. 3

   Use the fact that a square is never negative. $(x-1)^2\ge 0$ always, with equality precisely when $x-1=0$. So the strict inequality holds everywhere except that one value:

   $x\neq 1$

4. 4

   Write the answer as a union of two intervals.

   $(-\infty,1)\cup(1,\infty)$

   Test $x=1$: $1-2+1=0$, not $>0$ $\checkmark$. Test $x=0$: $1>0$ $\checkmark$. Test $x=1.01$: $0.0001>0$ $\checkmark$. Note how the answer would change with the sign: $\ge 0$ gives all reals, $<0$ gives the empty set, and $\le 0$ gives just $\{1\}$.