If x be real, then the minimum value of {x^2} | vedclass

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(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 mi

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$1$

Vedclass · Answer

(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 $1.$

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

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