To make sure the stuff under the square root is not negative, we force the (x+2) expression to be 0 or positive. Meaning that we write [tex]x+2 \ge 0[/tex] which solves to [tex]x \ge -2[/tex] when we subtract 2 from both sides
At the same time, the expression (5-x) under the other root is set to be greater than 0. We do not let (5-x) be equal to zero. Why not? Because this would cause a division by zero error. You cannot have the denominator be zero.
So, 5-x > 0 turns into 5 > x or x < 5 when we isolate x. This is after adding x to both sides.
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We have the two inequalities [tex]x \ge -2[/tex] and x < 5 at the same time. So we're describing some number between -2 and 5, where we include -2 but exclude 5. Again we're making sure x=5 is not allowed to avoid division by zero errors.
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Answer: Choice C) [tex]-2 \le x < 5[/tex]
