Gujarati
4-2.Quadratic Equations and Inequations
easy

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

A

$-1$

B

$0$

C

$1$

D

$2$

Solution

(c) ${x^2} – 8x + 17={x^2} – 8x + 16+1={(x-4)^2} + 1$
Since $x$ is real, so ${(x – 4)^2}$ is always positive and its least value is $0$ and so the minimum value of given expression is $1.$

Standard 11
Mathematics

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