Complex Numbers
Roots of polynomial equations
Grade 11
Question:
<p>If the equation \(z^4 + a_1 z^3 + a_2 z^2 + a_3 z + a_4 = 0\), where \(a_1, a_2, a_3, a_4\) are real coefficients different from zero, has a purely imaginary root, then the expression \(\dfrac{a_3}{(a_1 a_2)} + \dfrac{a_1 a_4}{(a_2 a_3)}\) has the value equal to</p>
<p>0</p>
<p>1</p>
<p>\(-2\)</p>
<p>2</p>
Step-by-Step Solution
Key Concept: If a polynomial with real coefficients has a purely imaginary root, then its negative conjugate is also a root. Using this property along with Vieta's formulas for the product and sum of roots allows us to establish relationships between the coefficients.
<p><strong>Step 1:</strong> Let the purely imaginary root be z = ib (where b ∈ ℝ, b ≠ 0). Since coefficients are real, -ib is also a root.</p><p><strong>Step 2:</strong> The polynomial can be factored as: P(z) = (z² + b²)(z² + pz + q), where the second factor has real coefficients.</p><p><strong>Step 3:</strong> Expanding: z⁴ + pz³ + qz² + b²z² + pb²z + qb² = z⁴ + pz³ + (q + b²)z² + pb²z + qb²</p><p><strong>Step 4:</strong> Comparing with z⁴ + a₁z³ + a₂z² + a₃z + a₄ = 0:<br/>• a₁ = p<br/>• a₂ = q + b²<br/>• a₃ = pb² = a₁b²<br/>• a₄ = qb² = (a₂ - b²)b²</p><p><strong>Step 5:</strong> From a₃ = a₁b²: b² = a₃/a₁</p><p><strong>Step 6:</strong> Calculate the given expression:<br/>a₃/(a₁a₂) + (a₁a₄)/(a₂a₃) = a₃/(a₁a₂) + (a₁·qb²)/(a₂·a₁b²) = a₃/(a₁a₂) + q/a₂</p><p><strong>Step 7:</strong> Since a₂ = q + b² = q + a₃/a₁, we have q = a₂ - a₃/a₁</p><p><strong>Step 8:</strong> Substituting: a₃/(a₁a₂) + (a₂ - a₃/a₁)/a₂ = a₃/(a₁a₂) + 1 - a₃/(a₁a₂) = 1</p><p><strong>∴ Answer:</strong> B</p>
Correct Answer: B