Algebra

Roots are sin(a) and cos(a). Sum = sin(a) + cos(a) = -b/a. Product = sin(a)cos(a) = c/a. Squaring the sum: sin^2(a) + cos^2(a) + 2sin(a)cos(a) = b^2/a^2. 1 + 2(c/a) = b^2/a^2. 1 + 2c/a = b^2/a^2. Multiply by a^2: a^2 + 2ac = b^2.

For a quadratic to have exactly one solution, the discriminant b^2 - 4ac must be 0. Here, 10^2 - 4 * a * 5 = 0, so 100 - 20a = 0, which means a = 5.

For roots to lie in (0, 5), the discriminant D = 16 - 4k must be >= 0 (so k <= 4). Also, f(0) > 0 (k > 0), f(5) > 0 (25 - 20 + k > 0 => k > -5), and the vertex 2 must be in (0, 5). Thus, k must be 1, 2, 3, 4.

Let roots be r1 and r2. r1 - r2 = 5. Sum of roots = b, product = c-5. (r1 - r2)^2 = (r1 + r2)^2 - 4r1r2. 5^2 = b^2 - 4(c-5). 25 = b^2 - 4c + 20. b^2 - 4c = 5, or b^2 = 4c + 5.

Let the roots be r1 and r2. r1*r2 = 1600. r1 >= 40. Since r1*r2 = 1600, r2 = 1600/r1. Since r1 <= 100, r2 >= 16. The equation is (x-r1)(x-r2) = x^2 - (r1+r2)x + r1*r2 = 0. So b = -(r1+r2) and c = 1600. We want to maximize b+c = 1600 - (r1+r2). To maximize this, we minimize r1+r2. Given r1*r2 = 1600 and r1 >= 40, the minimum sum occurs at the boundary r1=40, r2=40. Sum = 80. b+c = 1600 - 80 = 1520.

f(x) = x^2 - 10x + 23. Complete the square: (x-5)^2 - 25 + 23 = (x-5)^2 - 2. The minimum value of the function is -2, so y >= -2.

The minimum of f(x) = x^2 + ax + b occurs at x = -a/2. Given the vertex is (-1, 2), -a/2 = -1, so a = 2. Since f(-1) = 2, (-1)^2 + 2(-1) + b = 2, so 1 - 2 + b = 2, which means b = 3. Then |a - 2b| = |2 - 2(3)| = |2 - 6| = 4.

Solve |2x^2 - 3x - 5| = x + 1. Case 1: 2x^2 - 3x - 5 = x + 1 => 2x^2 - 4x - 6 = 0 => x^2 - 2x - 3 = 0 => (x-3)(x+1) = 0. Roots: 3, -1. Case 2: 2x^2 - 3x - 5 = -(x + 1) => 2x^2 - 2x - 4 = 0 => x^2 - x - 2 = 0 => (x-2)(x+1) = 0. Roots: 2, -1. Distinct roots are 3, -1, 2. Sum of squares = 3^2 + (-1)^2 + 2^2 = 9 + 1 + 4 = 14.

For x^2 - 10x + 22 = 0, m+n = 10 and mn = 22. m^2 + n^2 = (m+n)^2 - 2mn = 100 - 44 = 56. The value of (m^2 + n^2)(m + n) = 56 * 10 = 560.

For no real solutions, the discriminant D = b^2 - 4ac must be less than 0. Here a=3, c=5. D = b^2 - 4(3)(5) = b^2 - 60. So b^2 - 60 < 0, which means b^2 < 60.

Since f(x) >= 0 and f(2) = 0, (x - 2)^2 must be a factor. Let f(x) = a(x - 2)^2. Using f(4) = 6: a(4 - 2)^2 = 6, so 4a = 6, a = 1.5. Then f(-2) = 1.5 * (-2 - 2)^2 = 1.5 * 16 = 24.

No real roots for 2x^2+kx+5=0 means discriminant D = k^2 - 40 < 0, so k^2 < 40. k is in {-6, -5, ..., 5, 6}. Two distinct real roots for x^2+(k-5)x+1=0 means D = (k-5)^2 - 4 > 0, so (k-5)^2 > 4. This means k-5 > 2 or k-5 < -2, so k > 7 or k < 3. Combining k^2 < 40 and (k < 3 or k > 7): k can be -6, -5, -4, -3, -2, -1, 0, 1, 2. That is 9 values.

Given g(x) = 2x^2 - 2 and g(a) = 70, we solve 2a^2 - 2 = 70. This simplifies to 2a^2 = 72, so a^2 = 36, which means a = 6 or a = -6.