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Solve for real x: |x² + 3x + 2| + x + 1 = 0.
- x = -1 and x = -3
- x = -1 only
- x = -2 and x = -3
- No real solution
Correct answer: x = -1 and x = -3
Solution
Removing the absolute value requires casework on the sign of (x+1)(x+2). Each case gives a quadratic whose roots must be checked against the case's interval.
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