<p>The number of values of <em>a</em> for which equations \(x^3 + ax + 1 = 0\) and \(x^4 + ax^2 + 1 = 0\) have a common root is</p>
Step-by-Step Solution
Key Concept: If r is a common root of both equations, then r³ + ar + 1 = 0 and r⁴ + ar² + 1 = 0. Multiply the first equation by r to get r⁴ + ar² + r = 0, then subtract from the second equation to find r = 1, and use this to determine the possible values of a.
<p><strong>Step 1:</strong> Let r be a common root. Then:</p><p>r³ + ar + 1 = 0 ... (1)</p><p>r⁴ + ar² + 1 = 0 ... (2)</p><p><strong>Step 2:</strong> Multiply equation (1) by r:</p><p>r⁴ + ar² + r = 0 ... (3)</p><p><strong>Step 3:</strong> Subtract (3) from (2):</p><p>(r⁴ + ar² + 1) - (r⁴ + ar² + r) = 0</p><p>1 - r = 0</p><p>∴ r = 1</p><p><strong>Step 4:</strong> Substitute r = 1 into equation (1):</p><p>1 + a + 1 = 0</p><p>a = -2</p><p><strong>Step 5:</strong> Verify: For a = -2, check if x = 1 satisfies both equations:</p><p>x³ - 2x + 1 = 1 - 2 + 1 = 0 ✓</p><p>x⁴ - 2x² + 1 = 1 - 2 + 1 = 0 ✓</p><p><strong>Step 6:</strong> Check if there are other values by considering if r could be complex. The algebraic constraint r = 1 is independent of a's nature, so it's the only possibility.</p><p>∴ Answer: <strong>1</strong> (only a = -2)</p>
Correct Answer: B