Quadratic Equations
Roots in AP
Grade 11
Question:
<p><strong>For Problems 33 and 34</strong><br>The real numbers \(x_1, x_2, x_3\) satisfying the equation \(x^3 - x^2 + \beta x + \gamma = 0\) are in A.P.<br><br>All possible values of \(\gamma\) are</p>
<p>\(\left(-\frac{1}{9}, +\infty\right)\)</p>
<p>\(\left(-\frac{1}{27}, +\infty\right)\)</p>
<p>\(\left(\frac{2}{9}, +\infty\right)\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: If three roots are in A.P., express them as (a-d), a, (a+d). Use Vieta's formulas to establish relationships between coefficients and the roots, then eliminate variables to find the constraint on γ.
<p><strong>Step 1:</strong> Let the three roots in A.P. be (a-d), a, (a+d), where a is the middle term and d is the common difference.</p><p><strong>Step 2:</strong> Apply Vieta's formulas for x³ - x² + βx + γ = 0:</p><ul><li>Sum of roots: (a-d) + a + (a+d) = 1</li><li>This gives: 3a = 1, so a = 1/3</li></ul><p><strong>Step 3:</strong> Sum of products of roots taken two at a time:</p><p>(a-d)·a + a·(a+d) + (a-d)(a+d) = β</p><p>a² - ad + a² + ad + a² - d² = β</p><p>3a² - d² = β</p><p>Since a = 1/3: 3(1/9) - d² = β, so 1/3 - d² = β</p><p><strong>Step 4:</strong> Product of roots:</p><p>(a-d)·a·(a+d) = -γ</p><p>a(a² - d²) = -γ</p><p>(1/3)(1/9 - d²) = -γ</p><p>γ = -(1/3)(1/9 - d²) = d²/3 - 1/27</p><p><strong>Step 5:</strong> Since d² ≥ 0 for all real d, we have d² ∈ [0, ∞)</p><p>Therefore: γ = d²/3 - 1/27 ≥ 0 - 1/27</p><p>∴ γ ≥ -1/27 or γ ∈ [-1/27, ∞)</p><p><strong>All possible values of γ are: γ ≥ -1/27 (or γ ∈ [-1/27, ∞))</strong></p>
Correct Answer: B