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 $I$: $x^2 - x - 6 < 0$.$(x - 3)(x + 2) < 0$,which implies $-2 < x < 3$.In this case,the equation beco

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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 $-2 Case $II$: $x^2 - x - 6 \ge 0$.$(x - 3)(x + 2) \ge 0$,which implies $x \le -2$ 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 condition $x \le -2$ 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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