<p>The total number of integral values of \(a\) so that \(x^2 - (a+1)x + a - 1 = 0\) has integral roots is equal to</p>
Step-by-Step Solution
Key Concept: For a quadratic with integral roots, the discriminant must be a perfect square AND both roots must satisfy Vieta's formulas with integer constraints. Use the discriminant condition (a+1)² - 4(a-1) = a² - 2a + 5 to find when integer roots exist.
<p><strong>Step 1:</strong> Let roots be p and q (integers). By Vieta's formulas: p + q = a+1 and pq = a-1.</p><p><strong>Step 2:</strong> From these, p + q - pq = (a+1) - (a-1) = 2, so p + q - pq = 2, giving (1-p)(1-q) = -1.</p><p><strong>Step 3:</strong> Since p, q are integers, (1-p) and (1-q) are integers whose product is -1. The only factorizations of -1 are: 1×(-1) and (-1)×1.</p><p><strong>Step 4:</strong> Case 1: 1-p = 1 and 1-q = -1 gives p = 0, q = 2, so a-1 = 0, thus a = 1.</p><p><strong>Step 5:</strong> Case 2: 1-p = -1 and 1-q = 1 gives p = 2, q = 0, so a-1 = 0, thus a = 1.</p><p><strong>Step 6:</strong> Verify a = 1: x² - 2x + 0 = x(x-2) = 0 has roots 0 and 2 ✓. The discriminant is 4 - 0 = 4 (perfect square) ✓.</p><p><strong>Step 7:</strong> Check if discriminant D = a² - 2a + 5 = (a-1)² + 4 can be a perfect square for other values. We need (a-1)² + 4 = k² for integer k, so k² - (a-1)² = 4, giving (k - (a-1))(k + (a-1)) = 4.</p><p><strong>Step 8:</strong> Factor pairs of 4: (1,4), (2,2), (4,1), (-1,-4), (-2,-2), (-4,-1). Testing each yields a = 1 or a = 1 only.</p><p>∴ Total integral values of a = <strong>1</strong></p>
Correct Answer: B