Algebra

If 2 is a zero of x^2 + 3x + k, then f(2) = 0. 2^2 + 3(2) + k = 0. 4 + 6 + k = 0. 10 + k = 0, so k = -10.

Substitute x=2 into each equation. For C: 2(2^2) - 7(2) + 6 = 2(4) - 14 + 6 = 8 - 14 + 6 = 0. This is correct.

For real roots, the discriminant must be >= 0. For x^2 + rx + 64 = 0, r^2 - 4(64) >= 0 => r^2 >= 256 => r >= 16. For x^2 - 8x + r = 0, 64 - 4r >= 0 => r <= 16. Thus, r must be 16. Substituting r=16 into the options, only C works: 16^2 - 13(16) - 48 = 256 - 208 - 48 = 0.

Sum of roots = c + 6. Product of roots = 2(2c - 1) = 4c - 2. Given sum = 1/2 * product, so c + 6 = (4c - 2) / 2. c + 6 = 2c - 1, which gives c = 7.

A quadratic equation must be of the form ax^2 + bx + c = 0 where a is not 0. Option B: x + 3/x = x^2 becomes x^2 + 3 = x^3, which is a cubic equation, not quadratic.

For ax^2 + bx + c = 0 to have real roots, b^2 - 4ac must be nonnegative. Checking distinct choices from 1 through 5 gives 14 ordered triples satisfying this condition.

The first equation is x^2 + 10x + 17 = 0, roots a and b. Sum a+b = -10, product ab = 17. The second equation is x^2 + (a+b)x + ab - 8 = 0. Substituting values: x^2 - 10x + 17 - 8 = 0, so x^2 - 10x + 9 = 0. Factors are (x-1)(x-9) = 0. Roots are 1 and 9.

For roots to be imaginary, the discriminant D = b^2 - 4ac < 0. Here, (-k)^2 - 4(3)(12) < 0, so k^2 - 144 < 0, meaning k^2 < 144. This holds for any k in the interval (-12, 12). Since there are infinitely many real numbers in this interval, there are infinite values for k.

If x-1 is a factor, then f(1) = 0. 1^2 + k(1) + 9 = 0 => k = -10. Equation is x^2 - 10x + 9 = 0. Roots are (x-1)(x-9) = 0, so 1, 9.

For roots to be rational, the discriminant D = b^2 - 4ac must be a perfect square. D = (-7)^2 - 4(6)(k) = 49 - 24k. For 49 - 24k to be a perfect square, testing k=1 gives 25 (square), testing k=2 gives 1 (square).