The product of the roots of the cubic equation ax^3 + bx^2 + cx + d = 0 is given by -d/a. Here, d = -6 and a = 1, so the product is -(-6)/1 = 6.

By substituting x = 2 into the equation, we find that 2 is a root since 2^3 - 6(2^2) + 11(2) - 6 = 0.

The polynomial can be factored as (x-1)^3, indicating that it has one real root with multiplicity 3, hence all roots are real and equal.

The sum of the roots is given by -b/a = 3/1 = 3.

The polynomial can be factored as (x - 1)^3, indicating that all roots are real and equal.

The discriminant is given by b^2 - 4ac. Here, b = 2, a = 1, c = 1, so the discriminant is 2^2 - 4*1*1 = 0.

For the roots to be equal, the discriminant must be zero. Thus, 4^2 - 4*1*k = 0 leads to k = 4.

The product of the roots ab is given by c/a. Here, c = 6 and a = 1, so ab = 6.

For real and distinct roots, the discriminant must be greater than zero: 6^2 - 4*1*k > 0 leads to k < 9.

The discriminant must be greater than zero for real and distinct roots: (-4)^2 - 4*1*k > 0, which simplifies to 16 - 4k > 0, or k < 4.

For the roots to be equal, the discriminant must be zero. Thus, 2^2 - 4*1*k = 0 leads to k = 1.

The discriminant is 0, which indicates that the roots are real and equal.

For both roots to be positive, k must be greater than 1 (from Vieta's formulas).

Factoring the equation gives (x - 2)(x - 5) = 0, so the roots are 2 and 5.

The sum of the roots is -1 + (-2) = -3, and the product is (-1)(-2) = 2. Thus, k = 2.

The product of the roots of the equation ax^2 + bx + c = 0 is given by c/a. Here, c = 6 and a = 2, so the product is 6/2 = 3.

For equal roots, the discriminant must be zero. Thus, k^2 - 4*1*4 = 0, which gives k^2 = 16, so k = ±4.

The sum of the roots is 2 + 4 = 6, so k = 6.

The sum of the roots is 3 + 3 = 6, so k = 6.