<p>The quadratic equations \(x^2 - 6x + a = 0\) and \(x^2 - cx + 6 = 0\) have one root \(\alpha\) in common. The other roots of the first and second equations are integers in the ratio 4 : 3. Then equation \(x^2 - 3x - 4 = 0\) has</p>
<p>both roots more than <em>α</em></p>
<p>both roots less than <em>α</em></p>
<p>one root more than <em>α</em> and other less than <em>α</em></p>
<p>Can't say anything</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas on both equations with common root α: if the other roots are β and γ, then α + β = 6 and α + γ = c, and αβ = a, αγ = 6. The constraint that β:γ = 4:3 (both integers) creates a system to solve for α, a, and c.
<p><strong>Step 1:</strong> Let the other roots be β (from first equation) and γ (from second equation), with β:γ = 4:3. Let β = 4k and γ = 3k for integer k.</p><p><strong>Step 2:</strong> From Vieta's formulas on x² - 6x + a = 0: α + β = 6 and αβ = a. From x² - cx + 6 = 0: α + γ = c and αγ = 6.</p><p><strong>Step 3:</strong> Substituting: α + 4k = 6 → α = 6 - 4k, and α(3k) = 6 → α = 2/k.</p><p><strong>Step 4:</strong> Equating: 6 - 4k = 2/k → 6k - 4k² = 2 → 4k² - 6k + 2 = 0 → 2k² - 3k + 1 = 0 → (2k - 1)(k - 1) = 0.</p><p><strong>Step 5:</strong> So k = 1 or k = 1/2. Since β and γ must be integers, k = 1. Thus β = 4, γ = 3, and α = 2.</p><p><strong>Step 6:</strong> Therefore a = αβ = 2(4) = 8 and c = α + γ = 2 + 3 = 5. The equation x² - 3x - 4 = 0 factors as (x - 4)(x + 1) = 0, giving roots 4 and -1. These match β from the first equation and relate to the original constraint structure.</p><p>∴ Answer: C</p>
Correct Answer: C