<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, the discriminant must be a perfect square. The discriminant is (n+1)² - 4n(n+1) = (n+1)(1-3n), which must be a perfect square for integer n.
<p><strong>Step 1:</strong> For the equation nx² + (n+1)x + (n+1) = 0 to have rational roots, we need n ≠ 0 and discriminant Δ ≥ 0 and Δ must be a perfect square.</p><p><strong>Step 2:</strong> Calculate discriminant: Δ = (n+1)² - 4n(n+1) = (n+1)[(n+1) - 4n] = (n+1)(1-3n)</p><p><strong>Step 3:</strong> For real roots: (n+1)(1-3n) ≥ 0. This gives -1 ≤ n ≤ 1/3, so for integers: n ∈ {-1, 0}. Since n ≠ 0, only n = -1 is possible so far.</p><p><strong>Step 4:</strong> Check n = -1: Δ = (0)(-2) = 0 ✓ (perfect square). Equation becomes -x² + 0·x + 0 = 0, giving x = 0 (rational). ✓</p><p><strong>Step 5:</strong> For n > 1/3 or n < -1, the discriminant is negative (no real roots). For 1/3 < n < ∞ with n integer (n ≥ 1), check if (n+1)(1-3n) is a perfect square:</p><p>• n = 1: Δ = (2)(-2) = -4 < 0 ✗</p><p><strong>Step 6:</strong> Systematic check for small integer values shows only n = -1 yields both a quadratic equation and rational roots.</p><p>∴ Answer: <strong>1</strong> (only n = -1)</p>
Correct Answer: B