Quadratic Equations
Roots of polynomial equations
Grade 11
Question:
<p>If \(\alpha_1, \alpha_2, \alpha_3\) and \(\alpha_4\) are the roots of the equation \(x^4 + (2 - \sqrt{3})x^2 + (2 + \sqrt{3}) = 0\), then the value of \((1 - \alpha_1)(1 - \alpha_2)(1 - \alpha_3)(1 - \alpha_4)\) is equal to:</p>
<p>(a) 1</p>
<p>(b) 4</p>
<p>(c) \(2 + \sqrt{3}\)</p>
<p>(d) 5</p>
Step-by-Step Solution
Key Concept: Recognize this as a biquadratic equation in x². Substitute y = x² to convert it to a quadratic, then use the fact that (1 - α₁)(1 - α₂)(1 - α₃)(1 - α₄) = P(1) where P(x) is the original polynomial.
<p><strong>Step 1:</strong> Let y = x². The equation becomes y² + (2 - √3)y + (2 + √3) = 0.</p><p><strong>Step 2:</strong> For a polynomial P(x) = x⁴ + (2 - √3)x² + (2 + √3) with roots α₁, α₂, α₃, α₄, we have:</p><p>P(x) = (x - α₁)(x - α₂)(x - α₃)(x - α₄)</p><p><strong>Step 3:</strong> Therefore, P(1) = (1 - α₁)(1 - α₂)(1 - α₃)(1 - α₄)</p><p><strong>Step 4:</strong> Evaluate P(1):</p><p>P(1) = (1)⁴ + (2 - √3)(1)² + (2 + √3)</p><p>P(1) = 1 + (2 - √3) + (2 + √3)</p><p>P(1) = 1 + 2 - √3 + 2 + √3</p><p>P(1) = 5</p><p>∴ Answer: <strong>B (5)</strong></p>
Correct Answer: B