Question:

Step 1: Understand the requirement for two different pairs of equal roots. For the product of two quadratic factors to have two different pairs of equal roots, each quadratic must have a repeated root (discriminant = 0), and these two repeated roots must be different from each other. Step 2: Find values of $k$ for which the first quadratic has equal roots. For the first factor $x^2 + (2k-6)x + (7-3k) = 0$, we set the discriminant equal to zero: $$\Delta_1 = (2k-6)^2 - 4(1)(7-3k) = 0$$ Expanding: $$4k^2 - 24k + 36 - 28 + 12k = 0$$ $$4k^2 - 12k + 8 = 0$$ $$4(k^2 - 3k + 2) = 0$$ $$4(k-1)(k-2) = 0$$ Therefore: $k = 1$ or $k = 2$ Step 3: Find values of $k$ for which the second quadratic has equal roots. For the second factor $x^2 + (2k-2)x + (3k-5) = 0$, we set the discriminant equal to zero: $$\Delta_2 = (2k-2)^2 - 4(1)(3k-5) = 0$$ Expanding: $$4k^2 - 8k + 4 - 12k + 20 = 0$$ $$4k^2 - 20k + 24 = 0$$ $$4(k^2 - 5k + 6) = 0$$ $$4(k-2)(k-3) = 0$$ Therefore: $k = 2$ or $k = 3$ Step 4: Find the common value of $k$ where both quadratics have equal roots. For both discriminants to be zero simultaneously, we need the intersection of the two sets: - From Step 2: $\{1, 2\}$ - From Step 3: $\{2, 3\}$ The common value is: $k = 2$ Step 5: Verify that the two pairs of equal roots are different when $k = 2$. When $k = 2$: First factor: $x^2 + (2(2)-6)x + (7-3(2)) = 0$ $$x^2 - 2x + 1 = 0$$ $$(x-1)^2 = 0$$ This gives the repeated root: $x = 1$ Second factor: $x^2 + (2(2)-2)x + (3(2)-5) = 0$ $$x^2 + 2x + 1 = 0$$ $$(x+1)^2 = 0$$ This gives the repeated root: $x = -1$ Since $1 \neq -1$, the two pairs of equal roots are indeed different. ✓ Step 6: State the final answer. Only one value of $k$ (namely $k = 2$) satisfies the condition that the equation has two different pairs of equal roots. **The answer is Option 2: 1**