If |{x^2} - x - 6| = x + 2, then the values o | vedclass

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(b) $|{x^2} - x - 6| = x + 2$, then Case $I$ : ${x^2} - x - 6 < 0$ ==> $(x - 3)(x + 2) < 0$ ==> $ - 2 < x < 3$ In this c

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$-2, 2, 4$

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(b) $|{x^2} - x - 6| = x + 2$, then Case $I$ : ${x^2} - x - 6 < 0$ ==> $(x - 3)(x + 2) < 0$ ==> $ - 2 < x < 3$ In this case, the equation becomes ${x^2} - x - 6 = - x - 2$or ${x^2} - 4 = 0\,\,\, \Rightarrow x = \pm 2$ Clearly $x = 2$ satisfies the domain of the equation in this case. So $x = 2$ is a solution. Case $II$ : ${x^2} - x - 6 \ge 0.$So $x \le - 2$or $x \ge 3$ Then equation reduces to ${x^2} - x - 6 = 0 = x + 2$ i.e. ${x^2} - 2x - 8 = 0$or $x = - 2,\,4$ Both these values lies in the domain of the equation in this case, so $x = - 2,\,4$ are the roots. Hence roots are $x = - 2,\,2,\,4$.

If $|{x^2} - x - 6| = x + 2$, then the values of $x$ are

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