<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 independently have a discriminant of zero (repeated roots), and these roots must be different from each other.
<p><strong>Step 1:</strong> For two different pairs of equal roots in the product, each quadratic must have discriminant = 0.</p><p><strong>Step 2:</strong> For first quadratic: $x^2 + (2k-6)x + (7-3k) = 0$</p><p>Discriminant: $(2k-6)^2 - 4(7-3k) = 0$</p><p>$4k^2 - 24k + 36 - 28 + 12k = 0$</p><p>$4k^2 - 12k + 8 = 0$</p><p>$k^2 - 3k + 2 = 0$</p><p>$(k-1)(k-2) = 0$ → $k = 1$ or $k = 2$</p><p><strong>Step 3:</strong> For second quadratic: $x^2 + (2k-2)x + (3k-5) = 0$</p><p>Discriminant: $(2k-2)^2 - 4(3k-5) = 0$</p><p>$4k^2 - 8k + 4 - 12k + 20 = 0$</p><p>$4k^2 - 20k + 24 = 0$</p><p>$k^2 - 5k + 6 = 0$</p><p>$(k-2)(k-3) = 0$ → $k = 2$ or $k = 3$</p><p><strong>Step 4:</strong> Common solution from both: $k = 2$ makes both discriminants zero. But we need TWO DIFFERENT pairs of roots.</p><p><strong>Step 5:</strong> When $k=2$: First quad has double root at $x = \frac{-(2(2)-6)}{2} = 1$. Second quad has double root at $x = \frac{-(2(2)-2)}{2} = -1$. These are different! ✓</p><p><strong>Step 6:</strong> Check $k=1$ (first only) and $k=3$ (second only) don't work since we need BOTH quadratics with equal roots.</p><p>∴ Answer: <strong>1</strong> (only $k=2$)</p>
Correct Answer: C