If x is real, then value of {x^2} - 6x + 13 will not be less than

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4-2.Quadratic Equations and Inequations

easy

If $x$ is real, then the value of ${x^2} - 6x + 13$ will not be less than

A

$4$

B

$6$

C

$7$

D

$8$

Solution

(a) Let $y = {x^2} – 6x + 13 \Rightarrow {x^2} – 6x + 13 – y = 0$

Its discriminant $D \ge 0 \Rightarrow 36 – 4(13 – y) \ge 0$

==>$36 – 52 + 4y \ge 0 \Rightarrow 4y \ge 16 \Rightarrow y \ge 4$

Hence $y$ is not less than $4.$

Aliter : ${x^2} – 6x + 13 = {(x – 3)^2} + 4$

Obviously the minimum value is $4.$

Standard 11

Mathematics

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