Algebra

Roots of 8x^2 + 22x + 5 = 0 are x = (-22 +/- sqrt(484 - 160)) / 16 = (-22 +/- sqrt(324)) / 16 = (-22 +/- 18) / 16. Roots are -4/16 = -0.25 and -40/16 = -2.5. For tan^-1(x) to be real, x can be any real number. For sin^-1(x) or sec^-1(x), the domain is restricted (-1 to 1 for sin, |x| >= 1 for sec). Since -2.5 is outside the domain for sin^-1 and sec^-1, only tan^-1 is valid.

For distinct real roots, discriminant D > 0. D = 10^2 - 4(a)(a) = 100 - 4a^2. 100 - 4a^2 > 0 implies a^2 < 25, so -5 < a < 5. Among options, -4 is the only value in the range.

Let u = log2(x). The equation is x(3/4*u^2 + u - 5/4) = sqrt(2). This is a transcendental equation. Testing values or analyzing the function shows it has one irrational root.

For the quadratic equation (a^2+b^2)x^2 + 2(bc+ad)x + (c^2+d^2) = 0 to have real roots, the discriminant D must be >= 0. D = [2(bc+ad)]^2 - 4(a^2+b^2)(c^2+d^2) = 4(b^2c^2 + a^2d^2 + 2abcd) - 4(a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2) = 4(-a^2c^2 + 2abcd - b^2d^2) = -4(ac-bd)^2. For D >= 0, we must have ac=bd, which implies a/b = d/c, or ad=bc. This condition implies a^2, bd, c^2 are in GP if a^2/bd = bd/c^2, which is a^2c^2 = b^2d^2, consistent with ac=bd.

Given a+b+c=0, we can see that x=1 is a root of ax^2+bx+c=0 because a(1)^2+b(1)+c = a+b+c = 0. Since the product of the roots is c/a and one root is 1, the other root is c/a. Since a, b, c are rational, both roots are rational.

Since the coefficients are real, the complex conjugate 3-4i must also be a root. The sum of roots is 6 = -b/a and the product is (3+4i)(3-4i) = 9+16 = 25 = c/a. Substituting b = -6a and c = 25a into 31a+b+c gives 31a - 6a + 25a = 50a, which equals 2c (since 2c = 50a).

The discriminant of 4ax^2 + 3bx + 2c = 0 is D = (3b)^2 - 4(4a)(2c) = 9b^2 - 32ac. Given a+b+c=0, b=-(a+c). Substitute: 9(a+c)^2 - 32ac = 9(a^2 + 2ac + c^2) - 32ac = 9a^2 - 14ac + 9c^2. This quadratic in a/c has a negative discriminant (14^2 - 4*9*9 = 196 - 324 < 0), meaning it is always positive for real a, c. Thus D > 0, so roots are real and distinct.

Rationalize the root: (3 + sqrt(5)) / (2 - sqrt(5)) * (2 + sqrt(5)) / (2 + sqrt(5)) = (6 + 3*sqrt(5) + 2*sqrt(5) + 5) / (4 - 5) = (11 + 5*sqrt(5)) / -1 = -11 - 5*sqrt(5). For a quadratic with rational coefficients, the conjugate -11 + 5*sqrt(5) must also be a root. The sum of roots is -22 and the product is (-11)^2 - (5*sqrt(5))^2 = 121 - 125 = -4. The equation is x^2 - (sum)x + (product) = 0, which is x^2 + 22x - 4 = 0.

For 2x^2 + 2x + k = 0 to have distinct real roots, discriminant D = 4 - 8k > 0, so k < 0.5. For roots to be in [0, 1], let f(x) = 2x^2 + 2x + k. f(0) = k >= 0 and f(1) = 4 + k >= 0. Since k must be < 0.5 and k >= 0, roots would be negative. Thus, no such real k exists.

For x^2 + px + q = 0, the sum of roots is -p = tan(15) + tan(30) and the product is q = tan(15)tan(30). Using tan(15) = 2 - sqrt(3) and tan(30) = 1/sqrt(3), calculate p and q, then evaluate 2 + q - p.

Discriminant D = b^2 - 4ac = (-6)^2 - 4(5)(1) = 36 - 20 = 16.

The equation is 4x^2 - 2*sqrt(5)*x + 1 = 0. Roots are (2*sqrt(5) +/- sqrt(20 - 16)) / 8 = (2*sqrt(5) +/- 2) / 8 = (sqrt(5) +/- 1) / 4. These values correspond to sin(72) and sin(18), or sin(18) and cos(36).