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(C) The given equation is $|x^2+2x-8| = -(x-2) = 2-x$. Case $1$: $x^2+2x-8 \ge 0$. $(x+4)(x-2) \ge 0 \implies x \in (-\infty, -4] \cup [2, \infty)$. T

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$3$

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(C) The given equation is $|x^2+2x-8| = -(x-2) = 2-x$. Case $1$: $x^2+2x-8 \ge 0$. $(x+4)(x-2) \ge 0 \implies x \in (-\infty, -4] \cup [2, \infty)$. Then $x^2+2x-8 = 2-x \implies x^2+3x-10 = 0 \implies (x+5)(x-2) = 0$. So $x = -5$ and $x = 2$. Both satisfy the condition. Case $2$: $x^2+2x-8 $(x+4)(x-2) Then $-(x^2+2x-8) = 2-x \implies -x^2-2x+8 = 2-x \implies x^2+x-6 = 0 \implies (x+3)(x-2) = 0$. So $x = -3$ and $x = 2$. Since $x=2$ is not in $(-4, 2)$,we only take $x = -3$. The distinct real solutions are $\{-5, 2, -3\}$. Thus,there are $3$ distinct real solutions.

The number of distinct real solutions for the equation $|x^2+2x-8|+x-2=0$ is

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