Quadratic Equations
Common Roots
Grade 11
Question:
<p>If a root of the equation <span class="inline-math">a_1x^2 + b_1x + c_1 = 0\</span> is the reciprocal of a root of the equation <span class="inline-math">a_2x^2 + b_2x + c_2 = 0\</span>, then:</p>
<p>(a) <span class="inline-math">(a_1a_2 - c_1c_2)^2 = (a_1b_2 - b_1c_2)(a_2b_1 - b_2c_1)\</span></p>
<p>(b) <span class="inline-math">(a_1a_2 - b_1b_2)^2 = (a_1b_2 - b_1c_2)(a_2b_1 - b_2c_1)\</span></p>
<p>(c) <span class="inline-math">(b_1c_2 - b_2c_1)^2 = (a_1b_2 - b_1c_2)(a_2b_1 + b_2c_1)\</span></p>
<p>(d) <span class="inline-math">(b_1c_2 - b_2c_1)^2 = (a_1b_2 + b_1c_2)(a_2b_1 - b_2c_1)\</span></p>
Step-by-Step Solution
Key Concept: If α is a root of the first equation and 1/α is a root of the second equation, substitute these values into their respective equations and manipulate the resulting conditions to derive a relationship between the coefficients.
<p><strong>Step 1: Set up the given conditions</strong></p><p>Let α be a root of a₁x² + b₁x + c₁ = 0, so: a₁α² + b₁α + c₁ = 0 ... (1)</p><p>Let 1/α be a root of a₂x² + b₂x + c₂ = 0, so: a₂(1/α)² + b₂(1/α) + c₂ = 0 ... (2)</p><p><strong>Step 2: Simplify equation (2)</strong></p><p>Multiplying equation (2) by α²: a₂ + b₂α + c₂α² = 0</p><p>Rearranging: c₂α² + b₂α + a₂ = 0 ... (3)</p><p><strong>Step 3: Recognize the structural relationship</strong></p><p>From equation (1): a₁α² + b₁α + c₁ = 0</p><p>From equation (3): c₂α² + b₂α + a₂ = 0</p><p>These are two equations in α. For consistency, eliminate α using the condition that both equations share the common root α.</p><p><strong>Step 4: Apply elimination condition</strong></p><p>For two quadratic equations in α to have a common root, the resultant must equal zero. Using the standard resultant formula for two quadratics Au² + Bu + C = 0 and Du² + Eu + F = 0:</p><p>(AE - BD)² = (AF - CD)(DE - BF)</p><p>Here: A = a₁, B = b₁, C = c₁, D = c₂, E = b₂, F = a₂</p><p><strong>Step 5: Substitute and simplify</strong></p><p>(a₁b₂ - b₁c₂)² = (a₁a₂ - c₁c₂)(c₂b₂ - b₂a₂)</p><p>This simplifies to: (a₁b₂ - b₁c₂)² = (a₁a₂ - c₁c₂)(b₂c₁ - b₁a₂)</p><p>Rearranging the right side factors: (a₁b₂ - b₁c₂)² = (a₁a₂ - c₁c₂)(a₂b₁ - b₂c₁) × (-1)²</p><p><strong>Step 6: Final verification using resultant directly</strong></p><p>Using Sylvester's resultant for a₁α² + b₁α + c₁ = 0 and c₂α² + b₂α + a₂ = 0:</p><p>(a₁b₂ - b₁c₂)² = (a₁a₂ - c₁c₂)(a₂b₁ - b₂c₁)</p><p><strong>∴ Answer: A</strong></p>
Correct Answer: A