<p>The quadratic \(x^2 + ax + b + 1 = 0\) has roots which are positive integers, then \((a^2 + b^2)\) can be equal to</p>
Step-by-Step Solution
Key Concept: If a quadratic with integer coefficients has positive integer roots r and s, then a = -(r+s) and b+1 = rs, so b = rs-1. This constrains possible values of a² + b².
<p><strong>Step 1:</strong> Let the positive integer roots be r and s (where r, s ∈ ℤ⁺).</p><p><strong>Step 2:</strong> By Vieta's formulas: r + s = -a and rs = b + 1.</p><p>Therefore: a = -(r+s) and b = rs - 1.</p><p><strong>Step 3:</strong> Calculate a² + b²:</p><p>a² + b² = (r+s)² + (rs-1)²</p><p>= r² + 2rs + s² + r²s² - 2rs + 1</p><p>= r² + s² + r²s² + 1</p><p><strong>Step 4:</strong> Test small positive integer values:</p><p>• r=1, s=1: a² + b² = 1 + 1 + 1 + 1 = 4</p><p>• r=1, s=2: a² + b² = 1 + 4 + 4 + 1 = 10</p><p>• r=2, s=2: a² + b² = 4 + 4 + 16 + 1 = 25</p><p>• r=1, s=3: a² + b² = 1 + 9 + 9 + 1 = 20</p><p>• r=2, s=3: a² + b² = 4 + 9 + 36 + 1 = 50</p><p><strong>Step 5:</strong> The answer B matches one of these calculated values.</p><p>∴ Answer: B</p>
Correct Answer: B