Algebra

Subtracting the two equations: a(x^2-x^2) + 2c(x) - 2b(x) + b - c = 0. 2x(c-b) - (c-b) = 0. Since b != c, we can divide by (c-b) to get 2x - 1 = 0, so x = 1/2. Substituting x=1/2 into the first equation: a(1/4) + 2c(1/2) + b = 0. a/4 + c + b = 0. Multiplying by 4 gives a + 4c + 4b = 0.

Let f(x) = ax^2 + bx + c. Given 2a + 3b + 6c = 0, divide by 6 to get a/3 + b/2 + c = 0. This is equivalent to evaluating the integral of the quadratic or testing values. Specifically, f(0) = c and f(1) = a + b + c. Using the relation, one can show there is a root in (0, 1) by Rolle's theorem or intermediate value theorem on the antiderivative.

Roots alpha, beta satisfy alpha + beta = -p and alpha * beta = p^3. Given alpha = beta^2, we substitute to solve for p and then the roots.

For a pair of roots a and b, the transformed value is a/b + b/a = c^3/3 - 2, where c is the remaining root and abc = 3. Since c^3 + 2c - 3 = 0, this becomes -1 - 2c/3, so c = -3/2(1 + y).

Roots are 1+i, 1-i (conjugate pair) and 1+sqrt(2), 1-sqrt(2). Sum of roots = 4, product of roots = (1+1)(1-2) = -2. The quadratic factors are (x^2 - 2x + 2) and (x^2 - 2x - 1). Multiplying these: x^4 - 2x^3 - x^2 - 2x^3 + 4x^2 + 2x + 2x^2 - 4x - 2 = x^4 - 4x^3 + 5x^2 - 2x - 2 = 0.

For no real roots, discriminant b^2 - 4ac < 0. Option D is the incorrect statement based on the properties of the quadratic equation given.

For a quadratic f(x) = ax^2 + bx + c to be positive for all x, a must be > 0 and the discriminant must be < 0. f(x) + f(-x) = (ax^2 + bx + c) + (ax^2 - bx + c) = 2ax^2 + 2c. Since a > 0 and c > 0 (as f(0) > 0), this expression is always positive.

The roots of (x-a)(x-a-1)=0 are a and a+1. The roots of (x+a)(x+a^2-2)=0 are -a and 2-a^2. For the first roots to lie between the second, we need -a < a < a+1 < 2-a^2 or -a < a < a+1 < 2-a^2. Solving these inequalities leads to the specified interval. The resulting values for p and q yield q-p=2.

The second equation x^3 + 3x^2 + 3x + 2 = 0 can be written as (x+1)^3 + 1 = 0. The roots are x+1 = -1, -omega, -omega^2. So x = -2, -1-omega, -1-omega^2. If the quadratic has two common roots, it must be a factor of the cubic. The only way to have two common roots is if the quadratic is (x+1+omega)(x+1+omega^2) = x^2 + 2x + 1 + omega + omega^2 + omega^3 = x^2 + 2x + 1 - 1 + 1 = x^2 + 2x + 1. Comparing with ax^2+bx+c=0, a=b=c.

For a quadratic equation to remain unchanged when its roots are squared, the set of roots {a, b} must be identical to the set of squared roots {a^2, b^2}. This leads to two possibilities: either each root is equal to its own square (giving roots {0,0}, {1,1}, or {0,1}), or the roots are mapped to each other's squares (giving the complex cube roots of unity {w, w^2}). This results in exactly 4 such quadratic equations.

For zero to be a repeated root, both the constant term and the coefficient of x must be zero. The constant term gives lambda = 1 or 3, while the x coefficient is zero only when lambda = 1.

If one root is the reciprocal of the other, the product of roots (c/a) must be 1. Here, (2k^2 - k) / (5k) = 1. Assuming k is not 0, 2k^2 - k = 5k, so 2k^2 - 6k = 0, which gives 2k(k-3) = 0. Thus k=3.