JAMB 1997 · UME · Q13

Find the minimum value of $x^2 - 3x + 2$ for all real values of $x$ .

Worked solution (try it first)

Complete the square: half of $-3$ is $-\frac32$ , so $x^2 - 3x = \left(x - \frac32\right)^2 - \frac94$ .
So $x^2 - 3x + 2 = \left(x - \frac32\right)^2 - \frac94 + 2$
$= \left(x - \frac32\right)^2 - \frac14$ .
A square is never negative, so the minimum value is $-\frac14$ , option A.