Algebra

If 2 is a repeated root, f(2) = 0 and f'(2) = 0. f(x) = 3x^3 + px^2 + qx - 12. f'(x) = 9x^2 + 2px + q. f'(2) = 36 + 4p + q = 0 => 4p + q = -36. f(2) = 24 + 4p + 2q - 12 = 0 => 4p + 2q = -12. Subtracting: q = 24. Then 4p + 24 = -36 => 4p = -60 => p = -15. p + q = -15 + 24 = 9.

For x^4 - 4x^3 + ax^2 + bx + 1 = 0, let roots be r1, r2, r3, r4. Product = 1, Sum = 4. If all roots are 1, then (x-1)^4 = x^4 - 4x^3 + 6x^2 - 4x + 1 = 0. Thus a=6, b=-4.

For x^3+x+1=0, the product of (1+alpha)/(1-alpha) for roots alpha, beta, gamma is found by evaluating the function at 1 and -1. The result is 1/3.

The sum of the five cubic terms is a strictly increasing function of x, because its derivative is a sum of squared terms and is positive unless all constants are equal. Since the constants are not all equal, the equation crosses zero exactly once. Therefore, it has exactly one real root.

Using Cardano's method for x^3 + 12x - 12 = 0, let x = u + v. The solution involves cube roots. The expression 2*cbrt(2) - cbrt(4) is the correct real root.

Factoring the equation x^3 + 4x^2 - 4x - 16 = 0: x^2(x + 4) - 4(x + 4) = 0, which becomes (x^2 - 4)(x + 4) = 0. This factors further to (x - 2)(x + 2)(x + 4) = 0. The roots are 2, -2, and -4.

If x^3 = -8, then the roots a, b, c satisfy a^3 = -8, b^3 = -8, c^3 = -8. We want an equation with roots a^2, b^2, c^2. Let y = x^2, so x = y^(1/2). Substituting into x^3 = -8 gives (y^(1/2))^3 = -8, so y^(3/2) = -8. Squaring both sides gives y^3 = 64, or y^3 - 64 = 0.

By substituting the values from option C (x=3, y=4, z=1/2, u=1/3) into the equations: (3^2)(4^2)(1/2^2)(1/3) = 9*16*(1/4)(1/3) = 12. Checking others: (3^2)(4^2)(1/2)(1/3^2) = 9*16(1/2)(1/9) = 8. (1/2^2)(4)(3^2)(1/3^2) = (1/4)*4*9(1/9) = 1. 3(3)(4^2)(1/2^2)(1/3^2) = 3*3*16*(1/4)*(1/9) = 4. All equations hold.

Let sqrt(x-2) = y, so x-2 = y^2 or x = y^2 + 2. Substituting into the equation: y^2 + 2 + y = 4, which is y^2 + y - 2 = 0. Factoring gives (y+2)(y-1) = 0, so y = 1 or y = -2. Since y = sqrt(x-2) must be non-negative, y = 1. Thus, sqrt(x-2) = 1, x-2 = 1, x = 3. Checking x=3: 3 + sqrt(3-2) = 3+1 = 4. There is only one real root.

Let u = sqrt(x+10). Equation: u - 6/u = 5. u^2 - 5u - 6 = 0. (u-6)(u+1) = 0. u=6 or u=-1. Since u = sqrt(x+10) must be >= 0, u=6 is the only solution. sqrt(x+10) = 6 => x+10 = 36 => x=26. The root u=-1 is extraneous. Since u = sqrt(x+10) = -1 implies x+10 = 1, x = -9. -9 is between -10 and -6.