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

1

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Solution

The correct option is C $1$
Let $y = x^{2} - 8 x + 17 = (x^{2} - 2 \cdot 4 \cdot x + 4^{2}) + 1 = (x - 4)^{2} + 1$
We know that if $x$ is a real number then $x^{2} \geq 0$
$\Rightarrow (x - 4)^{2} \geq 0$
Hence minimum value of $y$ is $0 + 1 = 1$

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

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x^{2} - 8 x + 17

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

List-I	List-II
(A) Maximum value of $x y$ subject to $x + y = 7$ is	1) $72$
(B) If $l^{2} + m^{2} = 1$ , then the maximum value of $l + m$ is	2) $1$
(C) If $x + y = 12$ , then the minimum Value of $x^{2} + y^{2}$ is	3) $\sqrt{2}$
(D) Minimum value $x^{2} - 8 x + 17$ is	4) $\frac{49}{4}$
	5) $0$

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