Step-by-Step Solution
Key Concept: Absolute value equations require considering two cases: when the expression inside is non-negative and when it's negative. Critically, the RHS must be non-negative for the equation to have solutions, and solutions must be verified in the original equation.
<p><strong>Step 1:</strong> Note that for |x² + 4x + 3| = x + 1 to have solutions, we need x + 1 ≥ 0, so <strong>x ≥ -1</strong>.</p><p><strong>Step 2:</strong> Factor the expression: x² + 4x + 3 = (x + 1)(x + 3)</p><p><strong>Case 1 (when (x+1)(x+3) ≥ 0):</strong> This occurs when x ≤ -3 or x ≥ -1.</p><p>Combined with x ≥ -1, we need x ≥ -1. Then:<br/>(x + 1)(x + 3) = x + 1<br/>(x + 1)(x + 3) - (x + 1) = 0<br/>(x + 1)[(x + 3) - 1] = 0<br/>(x + 1)(x + 2) = 0</p><p>This gives x = -1 or x = -2. Since x ≥ -1, only <strong>x = -1</strong> is valid from this case.</p><p><strong>Case 2 (when (x+1)(x+3) < 0):</strong> This occurs when -3 < x < -1. Then:<br/>-(x + 1)(x + 3) = x + 1<br/>-(x² + 4x + 3) = x + 1<br/>-x² - 4x - 3 = x + 1<br/>-x² - 5x - 4 = 0<br/>x² + 5x + 4 = 0<br/>(x + 1)(x + 4) = 0</p><p>This gives x = -1 or x = -4. Since we need -3 < x < -1, neither solution lies in this interval. Note x = -1 is the boundary (excluded from this case).</p><p><strong>Step 3:</strong> Verify x = -1: |(-1)² + 4(-1) + 3| = |1 - 4 + 3| = |0| = 0, and (-1) + 1 = 0. ✓</p><p><strong>∴ Answer: x = -1</strong></p>
Correct Answer: x = -1