Quadratic Equations
Inequalities
Grade 11
Question:
<p>Solve for x, \((x^2 + 3x + 1)(x^2 + 3x - 3) \geq 5\)</p>
<p>(a) \((-\infty, -4] \cup [-2, -1] \cup [1, \infty)\)</p>
<p>(b) \((-\infty, -4] \cup [1, \infty)\)</p>
<p>(c) \((-\infty, -4] \cup [4, \infty)\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Substitute u = x² + 3x to convert this into a quadratic inequality in u, then solve (u + 1)(u - 3) ≥ 5, which simplifies to u² - 2u - 8 ≥ 0.
<p><strong>Step 1:</strong> Let u = x² + 3x. Then the inequality becomes:</p><p>(u + 1)(u - 3) ≥ 5</p><p><strong>Step 2:</strong> Expand:</p><p>u² - 3u + u - 3 ≥ 5</p><p>u² - 2u - 3 ≥ 5</p><p>u² - 2u - 8 ≥ 0</p><p><strong>Step 3:</strong> Factor the quadratic:</p><p>(u - 4)(u + 2) ≥ 0</p><p><strong>Step 4:</strong> Solve: u ≤ -2 or u ≥ 4</p><p><strong>Step 5:</strong> Substitute back u = x² + 3x:</p><p><strong>Case 1:</strong> x² + 3x ≤ -2 → x² + 3x + 2 ≤ 0 → (x + 1)(x + 2) ≤ 0 → -2 ≤ x ≤ -1</p><p><strong>Case 2:</strong> x² + 3x ≥ 4 → x² + 3x - 4 ≥ 0 → (x + 4)(x - 1) ≥ 0 → x ≤ -4 or x ≥ 1</p><p>∴ Answer: x ∈ [-2, -1] ∪ (-∞, -4] ∪ [1, ∞) or <strong>x ∈ (-∞, -4] ∪ [-2, -1] ∪ [1, ∞)</strong></p>
Correct Answer: A