Gujarati
4-2.Quadratic Equations and Inequations
easy

If $x$ be real, the least value of ${x^2} - 6x + 10$ is

A

$1$

B

$2$

C

$3$

D

$10$

Solution

(a) If ${x^2} – 6x + 10 = {(x – 3)^2} + 1$
For real $x$, least value of ${(x – 3)^2} + 1$is $1$.

Standard 11
Mathematics

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