Using the binomial theorem, the coefficient of x^2 in (2x - 3)^4 is given by 4C2 * (2)^2 * (-3)^2 = 6 * 4 * 9 = 216.

(1 + i)² = 1² + 2(1)(i) + i² = 1 + 2i - 1 = 2i.

Using the binomial theorem, the coefficient of x^4 in (a + b)^n is given by nCk * a^(n-k) * b^k. Here, n=6, a=x, b=-2, and k=2. Thus, the coefficient is 6C2 * (1)^4 * (-2)^2 = 15 * 4 = 60.

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

The discriminant D = b^2 - 4ac = 4^2 - 4(2)(2) = 16 - 16 = 0.

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

For equal roots, the discriminant must be zero: k^2 - 4*1*16 = 0. Solving gives k = -8.

The number of ways is 6C2 * 4C3 = 15 * 4 = 60.

The number of ways to choose 4 students from 10 is given by 10C4 = 210.

Using the identity a^2 + b^2 = (a + b)^2 - 2ab, we have 70 = 12^2 - 2ab. Thus, 70 = 144 - 2ab, leading to ab = 37.

a^2 = 4 and b^2 = 9, so 4 + 9 = 13.

log_2(x) + 2 = 6 implies log_2(x) = 4, so x = 2^4 = 16.

log_3(x) = 4 implies x = 3^4 = 81.

log_4(256) = log_4(4^4) = 4.

log_a(16) = 4 implies a^4 = 16. Since 16 = 2^4, we have a^4 = 2^4, thus a = 2.

log_b(25) = 2 implies b^2 = 25. Therefore, b = 5.

log_x(64) = 6 implies x^6 = 64. Since 64 = 2^6, we have x = 2.

The sum of the roots is 3 + (-1) = 2, hence p = -2.

By Vieta's formulas, the sum of the roots p + q = -b/a.

The equation can be factored as (x + 1)(x + 1) = 0, giving the root -1 with multiplicity 2.

Substituting x = -1 into the equation gives (-1)^2 + 2(-1) + k = 0, leading to 1 - 2 + k = 0, thus k = 1.

Using Vieta's formulas, p = -(3 + 4) = -7 and q = 3 * 4 = 12. Therefore, p + q = -7 + 12 = 5.

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.

The sum of the roots is -1 + -3 = -4, and the product is (-1)(-3) = 3. Thus, k = 3.

Using Vieta's formulas, k = (-1)(-2) = 2.

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.