I need help with this math question?

1.
kx(x - 1) = 2m - x
kx² - kx = 2m - x
kx² + (1 - k)x - 2m = 0

Sum of the roots of the equation above:
(p/2) + (q/2) = -(1 - k)/k
[(p/2) + (q/2)] * 2k = [-(1 - k)/k] * 2k
(p + q) k = 2k - 2
6k = 2k - 2
4k = -2
k = -1/2

Product of the roots of the equation above:
(p/2) * (q/2) = -2m/k
pq/4 = -2m/(-1/2)
3/4 = 4m
m = 3/16

Hence, k = -1/2 and m = 3/16

kx(x-1) = 2m-x can be written as x² +(1-k)/k * x -2m/k = 0. Do convince yourself of this.

Now, you are told that p/2 & q/2 are roots of this. That is, the above eqs can be written as (x - p/2)(x-q/2) = 0. Expanding this gives [left for you].

Answer that and we will take it from there if need be. Oh, and you must un-anon yourself too. 

kx * (x - 1) = 2m - x
kx^2 - kx = 2m - x
kx^2 + x - kx - 2m = 0
kx^2 + x * (1 - k) - 2m = 0

x = (k - 1 +/- sqrt(1 - 2k + k^2 + 8m * k)) / (2k)

p/2 = (k - 1 - sqrt(1 - 2k + 8mk + k^2)) / (2k)
q/2 = (k - 1 + sqrt(1 - 2k + 8mk + k^2)) / (2k)

p = (k - 1 - sqrt(1 - 2k + 8mk + k^2)) / k
q = (k - 1 + sqrt(1 - 2k + 8mk + k^2)) / k

p + q = 6

(k - 1 - sqrt(1 - 2k + 8mk + k^2)) / k + (k - 1 + sqrt(1 - 2k + 8mk + k^2)) / k = 6
(2k - 2) / k = 6
2 * (k - 1) / k = 6
(k - 1) / k = 3
k - 1 = 3k
k - 3k = 1
-2k = 1
k = -1/2

p * q = 3
(1/k^2) * ((k - 1)^2 - (1 - 2k + 8mk + k^2)) = 3
(k - 1)^2 - (1 - 2k + 8mk + k^2) = 3k^2

k = -1/2

(-1/2 - 1)^2 - (1 - 2 * (-1/2) + 8 * m * (-1/2) + (-1/2)^2) = 3 * (-1/2)^2
(-3/2)^2 - (1 + 1 - 4m + 1/4) = 3 * (1/4)
9/4 - (9/4 - 4m) = 3/4
6/4 = 9/4 - 4m
4m = 3/4
m = 3/16