Find the values of $x$ for which $x^{2} + 6 < 5 x$ .

2 < x < 3

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- 2 < x < 3

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- 3 < x < - 2

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- 2 < x

x > 3

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- 3 < x

x > 2

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Solution

The correct option is A $2 < x < 3$
Given, $x^{2} + 6 < 5 x$
$\Rightarrow x^{2} - 5 x + 6 < 0$
$\Rightarrow x^{2} - 5 x + 6 = (x - 2) (x - 3) < 0$
$\Rightarrow 2 < x < 3$

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