Algebra

The polynomial factors as (x + 2)(x + p), so its roots are -2 and -p. These roots are integers when p is an integer, and they are distinct when p is not 2.

Let u = |x-2|. Equation: u^2 + u - 2 = 0. (u+2)(u-1) = 0. u = 1 or u = -2. Since u is absolute value, u = 1. |x-2| = 1. x-2 = 1 or x-2 = -1. x = 3 or x = 1. Sum of roots = 3 + 1 = 4.

Using the limit as a approaches 0, we approximate (1+a)^n - 1 as n*a. The equation becomes (a/3)x^2 + (a/2)x + (a/6) = 0. Dividing by a/6 gives 2x^2 + 3x + 1 = 0. The roots are (2x+1)(x+1) = 0, so x = -1/2 and x = -1.

Since Sachin wrote the constant term incorrectly, his sum of roots (4 + 3 = 7) is correct, giving the coefficient of x as -7. Since Rahul wrote the coefficient of x incorrectly, his product of roots (3 * 2 = 6) is correct, giving the constant term as 6. The correct quadratic equation is x^2 - 7x + 6 = 0, which has roots 6 and 1.

For -x^2 + x - 2 = 0, multiply by -1 to get x^2 - x + 2 = 0. Using the quadratic formula x = (-b +/- sqrt(b^2 - 4ac)) / 2a, we get x = (1 +/- sqrt(1 - 8)) / 2 = (1 +/- sqrt(-7)) / 2 = (1 +/- i*sqrt(7)) / 2.

For roots to have opposite signs, the product of roots c/a must be negative. (a^2-4a)/2 < 0. a(a-4) < 0. This holds for 0 < a < 4.

For x^2 + 4x + c = 0 to have real roots, the discriminant D = 16 - 4c must be >= 0. So 16 >= 4c, or c <= 4. In the set {1, 2, 3, 4, 5, 6, 7, 8, 9}, the values of c satisfying this are {1, 2, 3, 4}. There are 4 such values out of 9, so the probability is 4/9.

For the quadratic equation Ax^2 + Bx + C = 0, roots are real and equal if the discriminant D = B^2 - 4AC = 0. Here, B^2 - 4AC = (2(ac+bd))^2 - 4(a^2+b^2)(c^2+d^2) = 4(a^2c^2 + b^2d^2 + 2acbd) - 4(a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2) = 4(2acbd - a^2d^2 - b^2c^2) = -4(ad-bc)^2. This is 0 only if ad=bc. Since the statement says they will be equal without qualification, it is generally false, but in many contexts, this is treated as a specific identity property.

For n^2 + 2sqrt(2)n + 1 = 0, the discriminant is (2sqrt(2))^2 - 4 = 4, which is positive, so the roots are real and distinct. The roots are -sqrt(2) + 1 and -sqrt(2) - 1, both irrational. Therefore, B is correct.

Given 4a+2b+c=0, so c = -4a-2b. The discriminant D = b^2 - 4ac = b^2 - 4a(-4a-2b) = b^2 + 16a^2 + 8ab. Since ab > 0, 8ab > 0. Also b^2 and 16a^2 are positive. Thus D > 0, implying real roots.