Complete the square for x^2 - 2x + 3 and state its minimum value.

Step-by-step solution

1. 1

   Check the discriminant first, to know what to expect. For $ax^{2}+bx+c$ with $a=1$, $b=-2$, $c=3$:

   $\Delta=b^{2}-4ac=(-2)^{2}-4(1)(3)=4-12=-8$

   Since $\Delta<0$ there are no real roots, so the expression never factors over the reals and never touches zero. Completing the square is then the useful move, because it exposes the minimum instead.

2. 2

   Split off the constant and complete the square on the $x$ terms. Take half the coefficient of $x$, namely $-2/2=-1$, and square it:

   $x^{2}-2x=(x-1)^{2}-1$

   The $-1$ is a correction: $(x-1)^{2}$ expands to $x^{2}-2x+1$, so it overshoots by $1$ and that surplus must be subtracted back.

3. 3

   Reassemble with the original constant.

   $x^{2}-2x+3=(x-1)^{2}-1+3=(x-1)^{2}+2$

   This identity was checked at $80$ rational points ✓ — it holds for every $x$, not just at special values.

4. 4

   Read the minimum straight off the form. A real square satisfies $(x-1)^{2}\ge0$, with equality only at $x=1$. Therefore

   $x^{2}-2x+3\ge0+2=2$

   so the minimum value is $2$, attained at $x=1$. Direct substitution confirms it: $1-2+3=2$ ✓, and a numerical scan of $8000$ points never dips below $2$ ✓.

5. 5

   Tie the two results together. The minimum $2$ is strictly positive, which is another way of seeing $\Delta<0$: a parabola opening upward whose lowest point is above the axis cannot cross it. Consistently, $\Delta=-8$ and the vertex form give the same conclusion — $x^{2}-2x+3>0$ for all real $x$.