$(x^{2} - x - 1) (x^{2} - x - 7) < - 5$
The range of values of $x$ satisfying the inequality is

(- 2, - 1) \cup (2, 3)

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(- 2, 1) \cup (2, 3)

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(2, 3)

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(- 2, - 1)

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Solution

The correct option is B $(- 2, - 1) \cup (2, 3)$
Let $x^{2} - x = t$
$(t - 1) (t - 7) + 5 < 0 \Rightarrow t^{2} - 8 t + 12 < 0$
$\Rightarrow (t - 6) (t - 2) < 0 \Rightarrow 2 < x^{2} - x < 6$
$x^{2} - x - 6 < 0$ & $x^{2} - x > 2$
$(x - 3) (x + 2) < 0$ & $(x - 2) (x + 1) > 0$
$\Rightarrow x \in (- 2, - 1) \cup (2, 3)$

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