Algebra

Sarika's roots (12, 16) imply the sum is 28, but she misread the x-coefficient, so the constant term is correct (12*16 = 192). Meera's roots (-22, -6) imply the product is 132, but she misread the constant, so the x-coefficient is correct (sum = -28, so coefficient is 28). The equation is x^2 - 28x + 192 = 0. Roots are 12 and 16.

For the roots of a quadratic equation ax^2 + bx + c = 0 to be reciprocals, the product of the roots (c/a) must equal 1. Here, k/3 = 1, which implies k = 3.

If 2 is a root of x^2 - 3x + k = 0, then substituting x = 2 must satisfy the equation: (2)^2 - 3(2) + k = 0. This simplifies to 4 - 6 + k = 0, which gives -2 + k = 0, so k = 2.

For roots a and b, (a+b)/2 = 8 implies a+b = 16. sqrt(ab) = 5 implies ab = 25. The quadratic equation is x^2 - (sum)x + (product) = 0, which is x^2 - 16x + 25 = 0.

If 2 is a root, 2(2^2) + k(2) + 4 = 0, so 8 + 2k + 4 = 0, 2k = -12, k = -6. The equation is 2x^2 - 6x + 4 = 0. Dividing by 2: x^2 - 3x + 2 = 0. (x-1)(x-2) = 0. Roots are 1 and 2. Both are positive integers.

x^2 + 2x + 1 = 0 is (x+1)^2 = 0. The roots are -1, -1. Since the roots are equal, the discriminant (b^2 - 4ac) = 2^2 - 4(1)(1) = 0.

Let f(x) = ax^4 + bx^3 + cx^2 + dx. Then f'(x) = 4ax^3 + 3bx^2 + 2cx + d. Given 27a + 9b + 3c + d = 0, this is f'(3) = 0. Also f(0) = 0 and f(3) = 81a + 27b + 9c + 3d = 3(27a + 9b + 3c + d) = 0. By Rolle's Theorem, there is a root of f'(x) in (0, 3).

For f(x) = x^2 - 8x + k, roots alpha and beta satisfy alpha+beta = 8 and alpha*beta = k. alpha^2 + beta^2 = (alpha+beta)^2 - 2*alpha*beta = 40. 8^2 - 2k = 40 => 64 - 2k = 40 => 2k = 24 => k = 12.

For a quadratic ax^2 + bx + c, the product of zeroes is c/a. Here, (2m - 1) / m = -1. So 2m - 1 = -m, which means 3m = 1, so m = 1/3.

Since x = 1 is a root for both, a(1)^2 + a(1) + 3 = 0 => 2a = -3 => a = -1.5. Also 1^2 + 1 + b = 0 => 2 + b = 0 => b = -2. Then ab = (-1.5)(-2) = 3.

Discriminant D = b^2 - 4ac. Here a = 3*sqrt(3), b = 10, c = sqrt(3). D = 10^2 - 4(3*sqrt(3))(sqrt(3)) = 100 - 4(3*3) = 100 - 36 = 64.

Discriminant D = b^2 - 4ac = (-2*sqrt(2))^2 - 4*(sqrt(3))(-2*sqrt(3)) = 8 - 4(-6) = 8 + 24 = 32. Since D > 0, roots are real and unequal.

For a quadratic equation Ax^2 + Bx + C = 0, the product of roots is C/A. Here, A = a(b-c), B = b(c-a), C = c(a-b). One root is 1, so the other root is C/A = c(a-b) / a(b-c).

The expression is (ac + ad + bc + bd)(a-c)(a-d)(b-c)(b-d) is not the right approach. The expression is symmetric. Using Vieta's formulas for the roots, the value evaluates to 16,000.

6|10r - 5| + 10 = 16 implies 6|10r - 5| = 6, so |10r - 5| = 1. This gives two cases: 10r - 5 = 1 (10r = 6, r = 0.6 or 3/5) and 10r - 5 = -1 (10r = 4, r = 0.4 or 2/5).