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 \(\beta\) are</p>
<p>\(\left(-\infty, \frac{1}{3}\right)\)</p>
<p>\(\left(-\infty, -\frac{1}{3}\right)\)</p>
<p>\(\left[\frac{1}{3}, \infty\right)\)</p>
<p>\(\left[-\frac{1}{3}, \infty\right)\)</p>
Step-by-Step Solution
Key Concept: If three roots are in A.P., use Vieta's formulas strategically: let roots be (a-d), a, (a+d). The sum of roots from Vieta's formulas directly gives you the middle term, which constrains the relationship between roots and coefficients.
<p><strong>Step 1:</strong> Let the three roots in A.P. be (a-d), a, (a+d).</p><p><strong>Step 2:</strong> By Vieta's formulas, sum of roots: (a-d) + a + (a+d) = 1</p><p>⟹ 3a = 1 ⟹ a = 1/3</p><p><strong>Step 3:</strong> The roots are (1/3 - d), 1/3, (1/3 + d).</p><p><strong>Step 4:</strong> By Vieta's formulas, sum of products of roots taken two at a time = β:</p><p>(1/3 - d)·(1/3) + (1/3)·(1/3 + d) + (1/3 - d)·(1/3 + d) = β</p><p><strong>Step 5:</strong> Simplify:</p><p>1/9 - d/3 + 1/9 + d/3 + (1/9 - d²) = β</p><p>2/9 + 1/9 - d² = β</p><p>1/3 - d² = β</p><p><strong>Step 6:</strong> Since d² ≥ 0 for all real d, we have β ≤ 1/3.</p><p>Also, by Vieta's formulas, product of roots = -γ, which can take any real value.</p><p>Therefore, all possible values of β are: <strong>β ≤ 1/3</strong> or <strong>β ∈ (-∞, 1/3]</strong></p><p>∴ Answer: A</p>
Correct Answer: A