(A) Given the quadratic expression $f(x) = x^{2}-8x+17$. To find the minimum value,we can complete the square: $f(x) = (x^{2}-8x+16) + 1$ $f(x) = (x-4

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(A) Given the quadratic expression $f(x) = x^{2}-8x+17$. To find the minimum value,we can complete the square: $f(x) = (x^{2}-8x+16) + 1$ $f(x) = (x-4)^{2} + 1$. Since $(x-4)^{2} \ge 0$ for all real $x$,the minimum value occurs when $(x-4)^{2} = 0$,which is at $x = 4$. Thus,the minimum value is $0 + 1 = 1$.

Class 11 Mathematics 4-2.Quadratic Equations and Inequations Solution of quadratic equations and Nature of roots

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If $x$ is real,then the minimum value of $x^{2}-8x+17$ is

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$1$
B
$2$
C
$3$
D
$4$

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If $x$ is real,then the minimum value of $x^{2}-8x+17$ is

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