The equation can be factored as (x - 3)^2 = 0, hence a = 3.

The equation can be factored as (2x - 3)(2x - 3) = 0, hence the roots are 3 and 3.

The sum of the roots is -1 + (-3) = -4, so -2 = -4, which is correct. The product of the roots is (-1)(-3) = 3, so k = 3.

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

For the roots to be equal, the discriminant must be zero. Thus, 4^2 - 4*1*k = 0 => 16 - 4k = 0 => k = 4.

Factoring gives (x - 3)(x - 7) = 0, so the roots are 3 and 7.

The discriminant must be greater than zero: (-2)^2 - 4*1*k > 0, which simplifies to k < 1.

For the roots to be equal, the discriminant must be zero. Thus, (-5)^2 - 4*1*k = 0 leads to 25 - 4k = 0, giving k = 6.25.

If the roots are both 2, then k = 2^2 - 4*2 = 4 - 8 = -4. Thus, k = 4.

Using Vieta's formulas, the sum of the roots is 7 and the product is k = 10.

The sum of the roots is given by -b/a = 8/2 = 4.

Using Vieta's formulas, the sum of the roots is -b/a = -12/3 = -4.

The 3rd term is given by 6C2 * a^4 * b^2 = 15 * a^4 * b^2 = 15ab^5.

The 3rd term is given by C(6, 2) * (x)^2 * (2)^4 = 15 * x^2 * 16 = 240x^2.

The 4th term is given by C(6,3) * (3x)^3 * (2)^3 = 20 * 27x^3 * 8 = 4320x^3.

The 5th term is given by C(7, 4)(2x)^4(-3)^3 = 35 * 16x^4 * (-27) = -1134x^5.

The 5th term is given by C(6, 4) * (3x)^4 * (-2)^2 = 15 * 81x^4 * 4 = 4860x^4.

The absolute value of -12 is 12.

The absolute value of -7 is 7.

The coefficient of x^0 in (x - 1)^5 is given by 5C5 * (-1)^5 = -1.

Using the binomial theorem, the coefficient of x^2 in (2x + 5)^4 is given by 4C2 * (2)^2 * (5)^2 = 6 * 4 * 25 = 600.

The coefficient of x^2 in (2x - 5)^5 is given by 5C2 * (2)^2 * (-5)^3 = 10 * 4 * (-125) = -5000.

The coefficient of x^2 in (3x + 4)^5 is C(5, 2) * (3)^2 * (4)^3 = 10 * 9 * 64 = 5760.

The coefficient of x^3 in (3x + 2)^5 is given by 5C3 * (3)^3 * (2)^2 = 10 * 27 * 4 = 1080.

The coefficient of x^3 in (x + 5)^6 is C(6, 3) * (5)^3 = 20 * 125 = 2500.

The coefficient of x^4 in (x + 3)^6 is C(6, 4) * 3^2 = 15 * 9 = 135.

The conjugate of z is given by z* = 7 + 4i.

The middle term in the expansion of (x + 2)^6 is the 4th term, which is C(6, 3)(x)^3(2)^3 = 20 * x^3 * 8 = 160x^3. The coefficient is 160.

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.

The product of the roots is given by c/a = 3/1 = 3.