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For what values of $x, x^{2} - 5 x + 6 is + v e$

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$2 < x < 3$

$x < 2 (o r) x > 3$

$x \in R$

$1 < x < 6$

answer is B.

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Detailed Solution

$f (x) = x^{2} - 5 x + 6 = x^{2} - 2 x - 3 x + 6$

$a = 1 > 0 = x (x - 2) - 3 (x - 2)$

$α < β = (x - 2) (x - 3)$

$\begin{array}{l} α = 2, β = 3 \\ (x - 2) (x - 3) > 0 \end{array}$

$x < 2 (o r) x > 3 \Rightarrow a > 0. E x p > 0$ is $+ v e$

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