Quadratic Equations
Roots of polynomial equations
Grade 11
Question:
<p>If \(\alpha, \beta, \gamma, \delta\) are the roots of the equation \(x^4 - Kx^3 + Kx^2 + Lx + M = 0\), where \(K, L,\) and \(M\) are real numbers, then the minimum value of \(\alpha^2 + \beta^2 + \gamma^2 + \delta^2\) is</p>
<p>(1) 0</p>
<p>(2) -1</p>
<p>(3) 1</p>
<p>(4) 2</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas to express the sum of squares of roots as (sum of roots)² - 2(sum of products of pairs), then apply the constraint that the coefficient of x³ and x² have a special relationship (both equal K) to minimize the expression.
<p><strong>Step 1:</strong> By Vieta's formulas for x⁴ - Kx³ + Kx² + Lx + M = 0:</p><p>• α + β + γ + δ = K</p><p>• αβ + αγ + αδ + βγ + βδ + γδ = K</p><p>• αβγ + αβδ + αγδ + βγδ = -L</p><p>• αβγδ = M</p><p><strong>Step 2:</strong> Calculate α² + β² + γ² + δ² using the identity:</p><p>α² + β² + γ² + δ² = (α + β + γ + δ)² - 2(αβ + αγ + αδ + βγ + βδ + γδ)</p><p>α² + β² + γ² + δ² = K² - 2K = K² - 2K</p><p><strong>Step 3:</strong> To find the minimum value, treat this as a function of K:</p><p>f(K) = K² - 2K</p><p>df/dK = 2K - 2 = 0 ⟹ K = 1</p><p><strong>Step 4:</strong> Minimum value = 1² - 2(1) = 1 - 2 = -1</p><p>Since this is achievable (e.g., when all roots satisfy the constraint), the minimum value is <strong>-1</strong>.</p><p>∴ Answer: A</p>
Correct Answer: A