Basic Mathematics & Logarithm
Modulus Equations
Grade 11

Question:

<p>Solve <br>\(|x^2 + x - 4| = |x^2 - 4| + |x|\)</p>
<p>\(x \in [-2, 0] \cup [2, \infty)\)</p>
<p>\(x \in (-\infty, -2] \cup [0, 2]\)</p>
<p>\(x \in [-2, 2]\)</p>
<p>\(x \in (-\infty, 0] \cup [2, \infty)\)</p>

Step-by-Step Solution

Key Concept: Analyze the equation by identifying critical points where expressions inside absolute values change sign (x = -2, 0, 2), then test each region to find which intervals satisfy the equation.
<p><strong>Step 1:</strong> Identify critical points where expressions change sign: x = -2, 0, 2 (from x² + x - 4 = 0, x = 0, and x² - 4 = 0)</p><p><strong>Step 2:</strong> Test region x < -2: |x² + x - 4| = x² + x - 4, |x² - 4| = x² - 4, |x| = -x<br>Equation: x² + x - 4 = x² - 4 - x → 2x = 0 → x = 0 (not in this region, invalid)</p><p><strong>Step 3:</strong> Test region -2 ≤ x < 0: |x² + x - 4| = -(x² + x - 4), |x² - 4| = 4 - x², |x| = -x<br>Equation: -x² - x + 4 = 4 - x² - x → 0 = 0 (identity: all x in [-2, 0) work)</p><p><strong>Step 4:</strong> Test region 0 ≤ x < 2: |x² + x - 4| = -(x² + x - 4), |x² - 4| = 4 - x², |x| = x<br>Equation: -x² - x + 4 = 4 - x² + x → -2x = 0 → x = 0 (boundary point, valid)</p><p><strong>Step 5:</strong> Test region x ≥ 2: |x² + x - 4| = x² + x - 4, |x² - 4| = x² - 4, |x| = x<br>Equation: x² + x - 4 = x² - 4 + x → 0 = 0 (identity: all x in [2, ∞) work)</p><p><strong>Step 6:</strong> Combine valid regions: x ∈ [-2, 0] ∪ [2, ∞)</p><p>∴ Answer: A</p>
Correct Answer: A

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