- Home
- Standard 11
- Mathematics
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