The set of real values of x satisfying the inequality | x 2 + x - 6 | 6 , is | x - 2 | + | x + 2 | 4 is

Question

The set of real values of $x$ satisfying the inequality

$| x^{2} + x - 6 |$ $> 6,$ is

$| x - 2 |$ $+$ $| x + 2 |$ $< 4$ is

Smart Achievers · Accepted Answer

(4)

$We have,$ $| x^{2} + x - 6 |$ $< 6$

$\Rightarrow$ $- 6 < x^{2} + x - 6 < 6$

$\Rightarrow$ $- 6 < x^{2} + x - 6 and x^{2} + x - 6 < 6$

$\Rightarrow$ $x^{2} + x > 0 and x^{2} + x - 12 < 0$

$\Rightarrow$ $x (x + 1) > 0 and (x + 4) (x - 3) < 0$

$\Rightarrow$ $x \in (- \infty, - 1) \cup (0, \infty) and - 4 < x < 3$

$\Rightarrow$ $x \in (- 4, - 1) \cup (0, 3)$

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