Algebra

For a quadratic equation to have equal roots, the discriminant (b^2 - 4ac) must be zero. For x^2 - kx + (k+2) = 0, the discriminant is (-k)^2 - 4(1)(k+2) = k^2 - 4k - 8 = 0. Using the quadratic formula, k = (4 +/- sqrt(16 - 4(1)(-8))) / 2 = (4 +/- sqrt(48)) / 2 = 2 +/- sqrt(12).

A quadratic has equal roots when its discriminant is zero. For ax^2 + bx + c, the discriminant is b^2 - 4ac, so it must equal 0.

Using the root formulas, the given condition simplifies to ab^2 + bc^2 = 2ca^2. This is exactly the condition that bc^2, ca^2, and ab^2 form an AP.

Since 1 is a root, a + b + c = 0, so c = -a - b. The discriminant of 4ax^2 + 3bx + 2c = 0 is 32a^2 + 32ab + 9b^2, which is positive for a nonzero. Therefore, the roots are real and unequal.

Let y = sqrt(x/(1-x)). The equation becomes y + 1/y = 13/6. Solving y^2 - (13/6)y + 1 = 0 gives y = 3/2 or 2/3. If sqrt(x/(1-x)) = 3/2, then x/(1-x) = 9/4, so 4x = 9 - 9x, 13x = 9, x = 9/13. If y = 2/3, x = 4/13.

The sum of coefficients is (b+c-2a) + (c+a-2b) + (a+b-2c) = 0. If the sum of coefficients is zero, one root is 1, and the other root is c/a (constant term / leading coefficient). Since a, b, c are rational, the roots are rational.

The roots of x^2 - 12x + 35 = 0 are 5 and 7. The roots of x^2 + 10x + 24 = 0 are -4 and -6. The positive roots are 5 and 7, and their sum is 12.

For a quadratic equation ax^2 + bx + c = 0, if the sum of coefficients (a + b + c) is 0, then x = 1 is a root. Here, (q-r) + (r-p) + (p-q) = 0. Thus, x = 1 is a root. The product of roots is c/a = (p-q)/(q-r). Since one root is 1, the other root is (p-q)/(q-r).