Algebra

For a quadratic ax^2+bx+c=0 with roots alpha and beta, the vertex of the parabola is at x = -b/(2a). If alpha < beta, the vertex lies between the roots.

Let f(x) = a_n*x^n + ... + a_1*x. Since f(alpha) = 0 and f(0) = 0, by Rolle's Theorem, there exists a root of f'(x) in (0, alpha). The derivative is f'(x) = n*a_n*x^(n-1) + ... + a_1. Thus, the root of the second equation is smaller than alpha.

Let f(x) = ax^2 + bx + c. We are given 2a + 3b + 6c = 0, which is f(2/6) * 6 = 0, or f(1/3) = 0. Since the function is continuous, if f(1/3) = 0, the root is exactly 1/3, which lies in the interval (0, 1).

Let f(x) = x^3 - 3x + k. For two roots in (0,1), we examine the derivative f'(x) = 3x^2 - 3. In (0,1), f'(x) < 0, so the function is strictly decreasing. A strictly decreasing function can have at most one root in an interval. Thus, it cannot have two different roots in (0,1).

Let f(x) = 4x^3 - 6x^2 + 12x + 9. f'(x) = 12x^2 - 12x + 12 = 12(x^2 - x + 1). The discriminant of x^2 - x + 1 is 1 - 4 = -3, so f'(x) is always positive. The function is strictly increasing and has exactly one real root.

Let f(t) = 2t^3 - 9t^2 + 30. For three real roots, the local maximum and minimum must straddle the value 'a'. f'(t) = 6t^2 - 18t = 6t(t-3). Critical points at t=0 and t=3. f(0) = 30 (local max), f(3) = 2(27) - 9(9) + 30 = 54 - 81 + 30 = 3 (local min). For three roots, 3 < a < 30.

3x = 20/7 - x. Add x to both sides: 4x = 20/7. Divide by 4: x = 5/7.

For the roots of z^2 + pz + q = 0 and the origin to form an equilateral triangle, the condition is p^2 = 3q.

For a quadratic equation ax^2 + bx + c = 0, the sum of zeroes is -b/a and the product is c/a. Here, -(-2)/k = -3k/k, which simplifies to 2/k = -3. Thus, k = -2/3.