If $x$ is real, the minimum value of $x^2 - 8x + 20$ is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 4
Given expression: $x^{2}-8x+20$
Complete the square:
$x^{2}-8x+20 = (x^{2}-8x+16)+4$
$=(x-4)^{2}+4$
Since $(x-4)^{2} \ge 0$, minimum occurs when $x=4$.
Minimum value $= 0 + 4 = 4$
The minimum value is 4.