Quadratic Equations
Roots of cubic equations
Grade 11
Question:
<p>If α, β and γ are the positive roots of the equation \(x^3 - px^2 + qx - 7 = 0\) such that \(\alpha\beta = 1\) and \(p, q \in R\) and \(p \leq 9\) then:</p>
<p>(a) \(|p + q| = 24\)</p>
<p>(b) \(p - q = -6\)</p>
<p>(c) \(\tan^{-1}\alpha + \tan^{-1}\gamma = \tan^{-1}\left(\dfrac{4}{3}\right)\)</p>
<p>(d) \(\tan^{-1}\alpha + \tan^{-1}\gamma = \tan^{-1}\left(\dfrac{4}{3}\right) = \pi\)</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas with the constraint αβ = 1 to express all roots in terms of one variable, then apply the product condition αβγ = 7 to find the roots, and finally use the sum constraint to establish the minimum value of p.
<p><strong>Step 1:</strong> Apply Vieta's formulas to x³ - px² + qx - 7 = 0:</p><ul><li>α + β + γ = p</li><li>αβ + βγ + γα = q</li><li>αβγ = 7</li></ul><p><strong>Step 2:</strong> Use constraint αβ = 1. From αβγ = 7, we get γ = 7.</p><p><strong>Step 3:</strong> Since αβ = 1 and β are positive roots, let α = t where t > 0, then β = 1/t.</p><p><strong>Step 4:</strong> Substitute into Vieta's sum:</p><p>p = α + β + γ = t + 1/t + 7</p><p><strong>Step 5:</strong> Find minimum of p. For t > 0, by AM-GM inequality:</p><p>t + 1/t ≥ 2√(t · 1/t) = 2</p><p>Equality holds when t = 1, so p_min = 2 + 7 = 9</p><p><strong>Step 6:</strong> Calculate q:</p><p>q = αβ + βγ + γα = 1 + γ(α + β) = 1 + 7(t + 1/t)</p><p><strong>Step 7:</strong> When p = 9 (t = 1): α = β = 1, γ = 7</p><p>This gives q = 1 + 7(2) = 15</p><p>∴ Statements B and C (involving relationships between p, q and the roots) are correct.</p>
Correct Answer: B,C