Solve the following inequality:
$(x^{2} - x - 1) (x^{2} - x - 7) < - 5$

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Solution

$(x^{2} - x - 1) (x^{2} - x - 7) < - 5$
Let $x^{2} - x$ be t
$(t - 1) (t - 7) < - 5$
$t^{2} - 8 t + 7 + 5 < 0 t^{2} - 8 t + 12 < 0 (t - 6) (t - 2) < 0 (x^{2} - x - 6) (x^{2} - x - 2) < 0 (x - 3) (x + 2) (x - 2) (x + 1) < 0 x \in (- 2, - 1) \cup (2, 3)$

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