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

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(B) Given equation: $|{x^2} - x - 6| = x + 2$Case $I$: ${x^2} - x - 6 < 0$$(x - 3)(x + 2) < 0 \Rightarrow -2 < x < 3$In this case,the equation becomes

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$-2, 2, 4$

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(B) Given equation: $|{x^2} - x - 6| = x + 2$Case $I$: ${x^2} - x - 6 $(x - 3)(x + 2) In this case,the equation becomes: $-(x^2 - x - 6) = x + 2$$-x^2 + x + 6 = x + 2$ $\Rightarrow x^2 = 4$ $\Rightarrow x = \pm 2$Since the domain is $-2 Case $II$: ${x^2} - x - 6 \ge 0$$(x - 3)(x + 2) \ge 0 \Rightarrow x \le -2$ or $x \ge 3$In this case,the equation becomes: $x^2 - x - 6 = x + 2$$x^2 - 2x - 8 = 0$ $\Rightarrow (x - 4)(x + 2) = 0$ $\Rightarrow x = 4$ or $x = -2$Both values satisfy the domain $x \le -2$ or $x \ge 3$.Combining both cases,the solutions are $x = -2, 2, 4$.

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

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