Algebra

The equation is x - 2/(x-1) = 1 - 2/(x-1). Adding 2/(x-1) to both sides gives x = 1. However, if x = 1, the term 2/(x-1) is undefined (division by zero). Therefore, there are no valid roots.

If 4 is a root of x^2 + ax + 8 = 0, then 16 + 4a + 8 = 0, so 4a = -24, a = -6. The second equation is x^2 - 6x + b = 0. For equal roots, the discriminant D = b^2 - 4ac = 0. Here, (-6)^2 - 4(1)(b) = 0, so 36 - 4b = 0, b = 9.

Since 1 is a root, substitute x = 1 into 2x^2 + ax + 3 = 0. This gives 2 + a + 3 = 0, so a = -5.

For equal roots, the discriminant D = b^2 - 4ac = 0. (3k)^2 - 4(4)(1) = 0. 9k^2 = 16. k^2 = 16/9. k = +/- 4/3.

For a quadratic f(x) = (a-5)x^2 - 2ax + (a-4), if one root is < 1 and the other > 2, then for a-5 > 0 (a > 5), f(1) < 0 and f(2) < 0. f(1) = a-5-2a+a-4 = -9 < 0 (always true). f(2) = (a-5)(4) - 4a + a - 4 = 4a - 20 - 4a + a - 4 = a - 24 < 0, so a < 24. Thus 5 < a < 24.

Let the roots be n and n+2. The sum of roots is 3/a = 2n+2 and the product is c/a = n(n+2). For real roots, the discriminant D = 9 - 4ac >= 0, so ac <= 9/4. Given the roots are positive odd integers, the smallest possible roots are 1 and 3, so 3/a = 4 implies a = 3/4. As roots increase, a decreases, leading to the interval (0, 3/4].