<p><strong>167.</strong> The number of values of \(k\) for which the equation \((x^2 + (2k-6)x + 7 - 3k)(x^2 + (2k-2)x + 3k - 5) = 0\) has two different pairs of equal roots, is equal to:</p>
Step-by-Step Solution
Key Concept: For the product of two quadratics to have two different pairs of equal roots, each quadratic must individually have equal roots (discriminant = 0), and the repeated roots from each quadratic must be different from each other.
<p><strong>Step 1:</strong> For <strong>two different pairs of equal roots</strong>, both quadratics must have discriminant = 0:</p><p>For <strong>Q₁:</strong> x² + (2k-6)x + (7-3k) = 0<br>Δ₁ = (2k-6)² - 4(7-3k) = 0<br>4k² - 24k + 36 - 28 + 12k = 0<br>4k² - 12k + 8 = 0<br>k² - 3k + 2 = 0<br>(k-1)(k-2) = 0 → <strong>k = 1 or k = 2</strong></p><p><strong>Step 2:</strong> For <strong>Q₂:</strong> x² + (2k-2)x + (3k-5) = 0<br>Δ₂ = (2k-2)² - 4(3k-5) = 0<br>4k² - 8k + 4 - 12k + 20 = 0<br>4k² - 20k + 24 = 0<br>k² - 5k + 6 = 0<br>(k-2)(k-3) = 0 → <strong>k = 2 or k = 3</strong></p><p><strong>Step 3:</strong> For both to have equal roots simultaneously: <strong>k = 2</strong> (common value)</p><p><strong>Step 4:</strong> Verify the roots are different when k = 2:<br>Q₁: x² - 2x + 1 = 0 → (x-1)² = 0 → root = 1 (repeated)<br>Q₂: x² + 2x + 1 = 0 → (x+1)² = 0 → root = -1 (repeated)<br>Since 1 ≠ -1, we have two different pairs of equal roots ✓</p><p>∴ Number of values of k = <strong>1</strong></p>
Correct Answer: C