Algebra

Solve |x^2 - x - 6| = x + 2. Case 1: x^2 - x - 6 = x + 2 => x^2 - 2x - 8 = 0 => (x-4)(x+2) = 0, so x=4, -2. Case 2: x^2 - x - 6 = -(x + 2) => x^2 - 4 = 0 => x = 2, -2. Combining these, the roots are -2, 2, 4.

The roots are 3 and -5. The x-coordinate of the vertex is the midpoint of the roots: (3 + -5) / 2 = -1. The y-coordinate f is d(-1 - 3)(-1 + 5) = d(-4)(4) = -16d.

If the roots of my^2 + ny + t = 0 are alpha and beta, the roots of the new equation are 1/alpha and 1/beta. Replacing y with 1/y in the original equation gives m(1/y)^2 + n(1/y) + t = 0. Multiplying by y^2 gives m + ny + ty^2 = 0, or ty^2 + ny + m = 0.

The roots p and q satisfy p + q = -v and pq = 15. Since p = m^8 + 324, and m is a non-negative integer, the smallest value for p is 324 (when m=0). Since p is a root of x^2 + vx + 15 = 0, p must be a factor of 15 (1, 3, 5, 15). None of these values are possible for p >= 324, so there are no such values for v.

For x^4 + 12x^3 + 46x^2 + ax + b to be a perfect square (x^2 + 6x + k)^2 = x^4 + 12x^3 + (36 + 2k)x^2 + 12kx + k^2. Matching coefficients: 36 + 2k = 46 => 2k = 10 => k = 5. Then a = 12k = 60 and b = k^2 = 25.

The roots of the first equation are 10sqrt(2) and 3sqrt(2). The roots of the second are sqrt(5) and -4sqrt(5), and every possible x value is greater than every possible y value, so x > y.

The product of the roots is r/p, which is positive, so the roots have the same sign. Their sum is -q/p, which is negative, so both distinct real roots must be negative.

Sum of roots = tan 19 + tan 26 = -p. Product of roots = tan 19 * tan 26 = q. Using tan(A+B) = (tan A + tan B) / (1 - tan A * tan B), tan 45 = 1 = (-p) / (1 - q). 1 = -p / (1 - q) => 1 - q = -p => p - q = -1 => q - p = 1.

For x^2 + 9|x| + 20 = 0, notice that x^2 is equivalent to |x|^2. The equation becomes |x|^2 + 9|x| + 20 = 0, which factors to (|x| + 4)(|x| + 5) = 0. Since |x| must be non-negative, |x| = -4 or |x| = -5 have no real solutions.

A^2 + 50A + 609 = 0. (A+21)(A+29) = 0. A = -21, -29. B^2 + 37B + 340 = 0. (B+17)(B+20) = 0. B = -17, -20. In all cases, A < B.

Solving the first quadratic equation 6A^2 - 29A + 35 = 0 gives the roots A = 2.5 and A = 2.33. Solving the second quadratic equation 10B^2 - 161B + 624 = 0 gives the roots B = 9.6 and B = 6.5. Since both values of A are strictly less than both values of B, we establish that A < B.

The equation is kx^2 - 2kx + 6 = 0. For equal roots, the discriminant D = b^2 - 4ac must be 0. Here, (-2k)^2 - 4(k)(6) = 0, which simplifies to 4k^2 - 24k = 0. Factoring gives 4k(k - 6) = 0. Since k cannot be 0 for a quadratic, k = 6.