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

From Vieta's formulas, p = -sum of roots = -5 and q = product of roots = 6.

The numbers are 4 and 8, as 4 + 8 = 12 and 4 × 8 = 32.

We know that x^2 + y^2 = (x + y)^2 - 2xy. Substituting the values, we get (10)^2 - 2(21) = 100 - 42 = 58.

|-3| = 3, so |x| + x = 3 - 3 = 0.

2x - 3 = 2(4) - 3 = 8 - 3 = 5.

|z|² = (1)² + (1)² = 1 + 1 = 2.

z^3 = (1 + i)^3 = 1 + 3i + 3i^2 + i^3 = 1 + 3i - 3 - i = -2 + 2i.

The conjugate of a complex number z = a + bi is given by z* = a - bi. Thus, the conjugate of z = 1 + i√3 is 1 - i√3.

z^2 = (1 + i√3)^2 = 1^2 + 2(1)(i√3) + (i√3)^2 = 1 + 2i√3 - 3 = -2 + 2i√3.

z² = (2 + 2i)² = 4 + 8i - 4 = 8i.

The modulus of z is given by |z| = √(3^2 + 4^2) = √(9 + 16) = √25 = 5.

z1 * z2 = (2 + 2i)(2 - 2i) = 4 - 4i^2 = 4 + 4 = 8.

The coefficient of x^3 in (2x - 3)^4 is given by 4C1 * (2)^3 * (-3)^1 = 4 * 8 * (-3) = -96. The coefficient is -108.

Using the binomial theorem, the coefficient of x^5 in (3x - 4)^7 is given by 7C5 * (3)^5 * (-4)^2 = 21 * 243 * 16 = 21 * 3888 = 81588.

Using the binomial theorem, the coefficient of x^4 in (x + 3)^6 is given by 6C4 * (3)^2 = 15 * 9 = 135.

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

The discriminant must be less than zero: (-4)^2 - 4*2*k < 0 leads to k > 8.

This is a perfect square trinomial: (x + 3)(x + 3) = 0.

For both roots to be negative, k must be greater than the square of half the coefficient of x, hence k > 9.

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

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.

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

The absolute value of -7 is 7.

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

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.