The solution set of $x^{2} - x - 6 < 0$ is

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(- 2, 3)

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[- 2, 3]

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

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Solution

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

case-1 : $x > 3$
$\Rightarrow (x - 3) (x + 2)$ is positive, no solution in $x > 3$

case-2 : $- 2 < x < 3$
$\Rightarrow (x - 3) (x + 2)$ is negative , all values are belong to solution that is $x \in (- 2, 3)$

case-3 : $x < - 2$
$\Rightarrow (x - 3) (x + 2)$ is positive, no solution in $x < - 2$

so answer is $x \in (- 2, 3)$

Suggest Corrections

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