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If $x$ is real, the minimum value of $x^2 - 8x + 20$ is |
4 2 0 -1 |
4 |
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. |
