Quadratic Equations
Nature of roots and inequalities
Grade 11
Question:
<p>Let <i>S</i> be the set of all non-zero real numbers such that the quadratic equation \(\alpha x^2 - x + \alpha = 0\) has two distinct real roots \(x_1\) and \(x_2\) satisfying the inequality \(|x_1 - x_2| < 1\). Which of the following intervals is(are) a subset(s) of <i>S</i>?</p>
<p>\(\left(-\dfrac{1}{2}, -\dfrac{1}{\sqrt{5}}\right)\)</p>
<p>\(\left(-\dfrac{1}{\sqrt{5}}, 0\right)\)</p>
<p>\(\left(0, \dfrac{1}{\sqrt{5}}\right)\)</p>
<p>\(\left(\dfrac{1}{\sqrt{5}}, \dfrac{1}{2}\right)\)</p>
Step-by-Step Solution
Key Concept: Use the discriminant condition for distinct real roots and express |x₁ - x₂| using Vieta's formulas: |x₁ - x₂| = √(Δ)/|a|. The inequality |x₁ - x₂| < 2 translates to a constraint on the discriminant relative to α².
<p><strong>Step 1:</strong> For two distinct real roots, discriminant Δ > 0:<br/>Δ = 1 - 4α² > 0 ⟹ α² < 1/4 ⟹ -1/2 < α < 1/2</p><p><strong>Step 2:</strong> By Vieta's formulas: x₁ + x₂ = 1/α and x₁x₂ = 1<br/>(x₁ - x₂)² = (x₁ + x₂)² - 4x₁x₂ = 1/α² - 4</p><p><strong>Step 3:</strong> Apply the condition |x₁ - x₂| < 2:<br/>|x₁ - x₂|² < 4 ⟹ |1/α² - 4| < 4</p><p><strong>Step 4:</strong> Since α² < 1/4, we have 1/α² > 4, so:<br/>1/α² - 4 < 4 ⟹ 1/α² < 8 ⟹ α² > 1/8 ⟹ |α| > 1/(2√2)</p><p><strong>Step 5:</strong> Combining constraints: α² ∈ (1/8, 1/4)<br/>Therefore: α ∈ (-1/2, -1/(2√2)) ∪ (1/(2√2), 1/2)<br/>∴ S = (-1/2, -√2/4) ∪ (√2/4, 1/2)</p>
Correct Answer: AD