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The minimum value of x2−8x+17x^2 - 8x + 17 on the set R\mathbb{R} of all real numbers is:

Solution

✅ Correct Option: 3

Complete the square: x2−8x+17=(x−4)2+1x^2 - 8x + 17 = (x-4)^2 + 1. Since (x−4)2≥0(x-4)^2 \geq 0, minimum value is 1 (achieved at x=4x = 4).

2022: 23 Aug Shift 1

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