Algebra

Since a, b, and c are in AP, we have 2b = a + c. Substituting the root x = 2 into the quadratic equation ax^2 + bx + c = 0 gives 4a + 2b + c = 0, which simplifies to 5a + 2c = 0, or c/a = -5/2. Since the product of the roots is c/a, the other root r satisfies 2 * r = -5/2, which gives r = -5/4.

Given |alpha - beta| = 1 and alpha/beta + beta/alpha = 5/2. The second equation is (alpha^2 + beta^2) / (alpha*beta) = 5/2. Using (alpha+beta)^2 = alpha^2+beta^2 + 2alpha*beta, we find alpha*beta = 2 and (alpha+beta)^2 = 9, so alpha+beta = 3. The equation is x^2 - 3x + 2 = 0.

For a quadratic equation with real coefficients, complex roots occur in conjugate pairs. If 3+i is a root, 3-i must also be a root. The sum of roots is (3+i) + (3-i) = 6. By Vieta's formulas, the sum of roots is -a. Thus, -a = 6, so a = -6.

For roots -1/4 and 1/2, their sum is 1/4 and their product is -1/8. Since the sum of roots is -b and the product is c, b = -1/4 and c = -1/8.

If roots are reciprocals, their product is 1. For ax^2 + bx + c = 0, the product of roots is c/a. Thus, c/a = 1, which implies a = c.

For x^2 + bx + ac = 0, roots are alpha and beta. For x^2 + ax + bc = 0, roots are alpha and gamma. Subtracting equations: (b-a)x + (ac-bc) = 0, so (b-a)x = c(b-a). Thus x = c. Since alpha is a root of both, alpha = c. Substituting alpha = c into the first equation: c^2 + bc + ac = 0, c(c+b+a) = 0. This implies alpha=c, beta=a, gamma=b.

alpha+beta = 2, alpha*beta = 4/3. Expression = (alpha^2+beta^2)/(alpha*beta) + 2(alpha+beta)/(alpha*beta) + 3(alpha*beta). = ((alpha+beta)^2 - 2*alpha*beta)/(alpha*beta) + 2(2)/(4/3) + 3(4/3). = (4 - 8/3)/(4/3) + 3 + 4 = (4/3)/(4/3) + 7 = 1 + 7 = 8.

If roots are alpha and beta, then alpha + beta = -b/a and alpha*beta = c/a. The new roots are (1-alpha)/alpha = 1/alpha - 1 and (1-beta)/beta = 1/beta - 1. Using the sum and product of these new roots for the equation px^2 + qx + r = 0, we find the relationship between coefficients.

For a quadratic equation ax^2 + bx + c = 0, the sum of roots a+b = -b/a and product ab = c/a. Here, a+b = -8 and ab = 4. The expression a/b + b/a simplifies to (a^2 + b^2) / ab, which is ((a+b)^2 - 2ab) / ab. Substituting values: ((-8)^2 - 2*4) / 4 = (64 - 8) / 4 = 56 / 4 = 14.

For x^2 + 5x - 7 = 0, a+b = -5 and ab = -7. The new roots are a+1 and b+1. Sum of new roots = a+b+2 = -5+2 = -3. Product of new roots = (a+1)(b+1) = ab + a+b + 1 = -7 - 5 + 1 = -11. The new equation is x^2 - (sum)x + (product) = x^2 + 3x - 11 = 0. Multiplying by 2 to match 2x^2 + px + q = 0 gives 2x^2 + 6x - 22 = 0. Thus p=6, q=-22. p+q = 6 - 22 = -16.

If -2 is a root, then (-2)^2 + 3(-2) + p = 0, so 4 - 6 + p = 0, which means p = 2. The equation is x^2 + 3x + 2 = 0. Factoring gives (x+1)(x+2) = 0, so the roots are -1 and -2.

Since a, b are roots of x^2 + px - q = 0, a^2 + pa - q = 0, so a^2 + pa = q. Since c, d are roots of x^2 + px + r = 0, c+d = -p and cd = r. (a-c)(a-d) = a^2 - a(c+d) + cd = a^2 - a(-p) + r = a^2 + pa + r. Since a^2 + pa = q, this equals q + r.

If a + ib and a - ib are roots of x^2 + ax + b = 0, the sum of roots is (a + ib) + (a - ib) = 2a. From the equation, the sum of roots is -a. Thus, 2a = -a, which implies 3a = 0, so a = 0.