<p>The number of integers <em>n</em> such that the equation \(nx^2 + (n+1)x + (n+1) = 0\) has only rational roots, is equal to:</p>
Step-by-Step Solution
Key Concept: For rational roots when n ≠ 0, the discriminant must be a perfect square. When n = 0, we get a linear equation. Both conditions together severely restrict possible integer values of n.
<p><strong>Step 1:</strong> Handle the case n = 0 separately.</p><p>When n = 0: x + 1 = 0, so x = -1 (rational). ✓</p><p><strong>Step 2:</strong> For n ≠ 0, use the discriminant condition.</p><p>Discriminant: Δ = (n+1)² - 4n(n+1) = (n+1)[(n+1) - 4n] = (n+1)(1-3n)</p><p><strong>Step 3:</strong> For rational roots, Δ must be a perfect square ≥ 0.</p><p>We need: (n+1)(1-3n) = k² for some integer k ≥ 0</p><p><strong>Step 4:</strong> Check integer values systematically.</p><p>• n = 1: Δ = (2)(-2) = -4 < 0 ✗</p><p>• n = -1: Δ = (0)(4) = 0 ✓ (gives x = 1 with multiplicity 2)</p><p>• n = 2: Δ = (3)(-5) = -15 < 0 ✗</p><p>• n = -2: Δ = (-1)(7) = -7 < 0 ✗</p><p><strong>Step 5:</strong> Observe that for n ≥ 1, we have 1 - 3n < 0 and n + 1 > 0, so Δ < 0.</p><p>For n ≤ -2, we have (n+1)(1-3n) becomes negative or fails to be a perfect square when checked individually.</p><p>Only n = 0 and n = -1 work.</p><p>∴ Answer: <strong>2</strong> (Option B)</p>
Correct Answer: B